Index Part2

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Definition of Logarithm

$a^{x}=N\Leftrightarrow$ $\log_{a}N=x\,$

Note : It is VERY important to know how to change from index form to log form and vice versa

- $\log\,$
- $\log_{{\color{Red}{a}}}$ (base)
- $\log_{{\color{Red}{a}}}\quad={\color{Blue}{x}}$ (index)
- $\log_{{\color{Red}{a}}}N$ (Number) $={\color{Blue}{x}}$

Value of a & N

$a\,$ must be a positive value
Thus, $N\,$

Special notation

$\lg x\,$ $=\log_{10}x\,$
$\ln x\,$ $=\log_{e}x\,$
These two should just be considered as short form rather than actual formulas

Examples

$\log_{2}{4}=\,$ $2\,$ Note that we are essentially asking $2^{?}=4\,$
$\log_{2}{8}=\,$ $3\,$
$\log_{3}{\frac{1}{3}}=\,$ $-1\,$

Exercise 1

Complete the following table

1	$2^{3}=8\,$	$\log_{2}8=3\,$
2	$\frac{1}{9}=3^{-2}$	$\log_{3}\frac{1}{9}=-2\,$
3	$10^{2}=100\,$	$\log_{10}100=2\,$ or $\lg 100=2\,$
4	$10^{x}=y\,$	$\log_{10}y=x\,$ or $\lg y=x\,$
5	$e^{a}=b\,$	$\log_{e}b=a\,$ or $\ln b=a\,$
6	$e^{0}=1\,$	$\log_{e}1=0\,$ or $\ln 1=0\,$
7	$2^{5}=x\,$	$\log_{2}x=5\,$
8	$x^{5}=2\,$	$\log_{x}2=5\,$
9	$10^{y}=x\,$	$\lg x=y\,$
10	$e^{x}=y\,$	$\ln y=x\,$

Laws of Logarithm

a) $\log_{a}xy\,$ $=\log_{a}x+\log_{a}y\,$

Prove
- $\log_{a}\underbrace{xy}_{{\color{Blue}\mbox{times}}}=\log_{a}x\underbrace{+}_{{\color{Blue}\mbox{plus}}}\log_{a}y\,$
- $a^{m}{\color{Blue}\times} a^{n}=a^{m{\color{Blue}+}n}\,$
- Comparing ${\color{Blue}a^{m}\times{a}^{n}}=a^{m+n}$ with $\log_{a}{\color{Blue}xy}=\log_{a}x+\log_{a}y$
- We now know where to start
- $\mbox{Let } x=a^{m}, y=a^{n}\,$ We see that we will need $xy\,$
- $xy=\,$ $a^{m}\times a^{n}$ Which according to index laws...
- $=a^{m+n}\,$
- $\log_{\color{Red}a}xy={{\color{Blue}m+n}}$
- $\mbox{Let } x=a^{m}, y=a^{n}\,$
- $\therefore \log_{a}x=m, \log_{a}x=n$
- $\therefore \log_{a}xy=\log_{a}x+\log_{a}y$
- $\begin{array}{ll} & \mbox{Let } x=a^{m}, y=a^{n} \\ & \therefore \log_{a}x=m, \log_{a}x=n \\ & xy = a^{m}\times a^{n} \\ & \quad = a^{m+n} \\ & \therefore \log_{a}xy=m+n\\ & \therefore \log_{a}xy=\log_{a}x+\log_{a}y\\ \end{array}$

b) $\log_{a}\left(\frac{x}{y}\right)\,$ $=\log_{a}x-\log_{a}y\,$

Try it on your own first
- $\begin{align} & \mbox{Let } x=a^{m}, y=a^{n} \\ & \therefore \log_{a}x=m, \log_{a}x=n \\ & \frac{x}{y} = \frac{a^{m}}{a^{n}} \\ & \quad = a^{m-n} \\ & \therefore \log_{a}\left(\frac{x}{y}\right)=m-n\\ & \therefore \log_{a}\left(\frac{x}{y}\right)=\log_{a}x-\log_{a}y\\ \end{align}$

c) $\log_{a}x^{n}\,$ $=n\log_{a}x\,$

- $\mbox{Let } a^{m}=x\, \therefore \log_{a}x=m$ We need $x^{n}\,$
- $x^{n}=\left(a^{m}\right)^{n}$
- $\log_{a}x^{n}=mn\,$
- $\log_{a}x^{n}=\left(\log_{a}x\right)n$
- $=n\log_{a}x\,$

d) $\log_{a}a\,$ $=1\,$

$a^?=a\,$ $a^{1}=a\,$

e) $\log_{a}1\,$ $=0\,$

$a^?=1\,$ $a^{0}=1\,$

Note : All the laws apply for $\ln\,$ / $\lg\,$

Common Mistakes

$\log_{a}\left(x+y\right)=$
$\left(\log_{a}x\right)\left(\log_{a}y\right)=$
$\left(\log_{a}x\right)^{n}=$
$\log_{a}xy^{n}=n\log_{a}xy\,$ Correct? WRONG, the $n\,$ is only for $y\,$

Exercise 2

Complete the following table

1	$\lg x^{2}y\,$	$2\lg x + \lg y\,$
2	$\lg \frac{1}{\left(xy\right)^{2}}\,$	$-2\lg x -2 \lg y\,$
3	$\lg \sqrt{\frac{xy^{3}}{z^{2}}}$	$\frac{1}{2}\lg x+\frac{3}{2}\lg y-\lg z$
4	$\lg \left(\frac{x^{2}}{y^{3}}\right)$	$2\lg x-3\lg y\,$
5	$\lg \sqrt{xy^{3}}$	$\frac{1}{2}\lg x+\frac{3}{2}\lg y\,$
6	$\lg \frac{1}{xy^{2}}$	$-\lg x-2\lg y\,$

Change of base

f) $\log_{a}b\,$ $=\frac{\log_{c}b}{\log_{c}a}$ (change of base to $c\,$ )

Prove : (the prove is important and can be asked in STPM)
How do we even manipulate ?
- $\mbox{Let } \log_{a}b=x\,$ Change this to index form...
- $x\log_{c}a=\log_{c}b\,$
- $x=\frac{\log_{c}b}{\log_{c}a}$
- $\therefore \log_{a}b=\frac{\log_{c}b}{\log_{c}a}$

g) $\log_{a}b\,$ $=\frac{1}{\log_{b}a}$ (change of base to $b\,$ )

Prove (We can either used a similar method, or if the above is already proved, we can use it)
$\begin{align} & \mbox{Let } \log_{a}b=x \\ & \therefore a^{x}=b \\ & \therefore \log_{b}a^{x}=\log_{b}b \\ & x\log_{b}a=1 \\ & x=\frac{1}{\log_{b}a}\\ & \therefore \log_{a}b = \frac{1}{\log_{b}a} \end{align}$ OR $\begin{align} & \log_{a}b=\frac{\log_{c}b}{\log_{c}a} \\ & \mbox{If } c=b, \\ & \therefore \log_{a}b=\frac{\log_{b}b}{\log_{b}a}\\ & \qquad \qquad =\frac{1}{\log_{b}a} \end{align}$

Examples

$\log_{4}{8}\,$ $=\frac{3}{2}$
Since it is harder (but not impossible) to see it from , we change the base to , since both numbers are base
- $\log_{4}{8}\,$
- $={\color{Red}\frac{\log_{2}\quad}{\log_{2}\quad}}$ $\frac{\log_{2}{\color{Red}8}}{\log_{2}{\color{Red}4}}$
- $=\frac{3}{2}$

$\log_{9}{3}\,$ $=\frac{1}{2}$
We directly interchange the base and the number
- $=\frac{1}{\log_{3}{9}}=\frac{1}{2}$
$\log_{\frac{1}{3}}\sqrt{3}$ $=\frac{\log_{3}{\sqrt{3}}}{\log_3{\frac{1}{3}}}=\frac{\frac{1}{2}}{-1}=-\frac{1}{2}$

Algebra (Revision)

Just something I find some students can get confused/careless at times

$\frac{a}{\frac{b}{c}}=$

$\frac{a}{\left(\frac{b}{c}\right)}=$ $\frac{ac}{b}$

$\frac{\left(\frac{a}{b}\right)}{c}=$ $\frac{a}{bc}$

$\frac{\left(\frac{2}{3}\right)}{x}=$ $\frac{2}{3x}$

$\frac{2}{\left(\frac{3}{x}\right)}=$ $\frac{2x}{3}$

$\frac{\left(\frac{x}{2}\right)}{3}=$ $\frac{x}{6}$

$\frac{2}{\left(\frac{x}{3}\right)}=$ $\frac{6}{x}$

$\frac{\left(\frac{3}{x}\right)}{2}=$ $\frac{3}{2x}$

$\frac{x}{\left(\frac{2}{3}\right)}=$ $\frac{3x}{2}$

Exercise 3

Express the following in terms of $u=\log_{2}x\,$

1) $\log_{2}x^{3}\,$ $=3u\,$

- $\begin{align} \log_{2}x^{3} & = 3\log_{2}x\\ & =3u \end{align}$

2) $\log_{2}\frac{1}{\sqrt[3]{x^{2}}}\,$ $=-\frac{2}{3}u\,$

- $\begin{align} \log_{2}\frac{1}{\sqrt[3]{x^{2}}} & = \log_{2}x^{-\frac{2}{3}}\\ & = -\frac{2}{3}\log_{2}x\\ & = -\frac{2}{3}u \end{align}$

3) $\log_{2}2x^{3}\,$ $=1+3u\,$

- $\begin{align} \log_2{2}x^{3} & = \log_{2}2+3\log_{2}x\\ & =1+3u \end{align}$

4) $\log_{2}\frac{x}{4}\,$ $=u-2\,$

- $\begin{align} \log_{2}\frac{x}{4} & = \log_{2}x-\log_{2}4\\ & =u-2 \end{align}$

5) $\log_{2}\frac{1}{8x}\,$ $=-3-u\,$

- $\begin{align} \log_{2}\frac{1}{8x} & = -\log_{2}8x\\ & =-\left(\log_{2}8+\log_{2}x\right)\\ & =-3-u \end{align}$

6) $\log_{2}4\sqrt {x}\,$ $=2+\frac{1}{2}u\,$

- $\begin{align} \log_{2}4\sqrt {x} & = \log_{2}4+\frac{1}{2}\log_{2}{x}\\ & =2+\frac{1}{2}u \end{align}$

7) $\log_{4}x\,$ $=\frac{1}{2}u\,$

- $\begin{align} \log_{4}x & = \frac{\log_{2}x}{\log_{2}4}\\ & =\frac{1}{2}u \end{align}$

8) $\log_{\frac{1}{2}}x\,$ $=-u\,$

- $\begin{align} \log_{\frac{1}{2}}x & = \frac{\log_{2}x}{\log_{2}\frac{1}{2}}\\ & =-u \end{align}$

9) $\log_{\sqrt{8}}x\,$ $=\frac{2}{3}u\,$

- $\begin{align} \log_{\sqrt{8}}x & = \frac{\log_{2}x}{\log_{2}\sqrt{8}}\\ & = \frac{\log_{2}x}{\frac{1}{2}\log_{2}{8}}\\ & =\frac{u}{\left(\frac{3}{2}\right)}\\ & =\frac{2}{3}u \end{align}$

10) $\log_{\frac{1}{\sqrt{8}}}x\,$ $==-\frac{2}{3}u\,$

- $\begin{align} \log_{\frac{1}{\sqrt{8}}}x & = \frac{\log_{2}x}{\log_{2}\frac{1}{\sqrt{8}}}\\ & = \frac{\log_{2}x}{-\frac{1}{2}\log_{2}{8}}\\ & =\frac{u}{\left(-\frac{3}{2}\right)}\\ & =-\frac{2}{3}u \end{align}$

11) $\log_{4}2x\,$ $=\frac{1+u}{2}$

- $\begin{align} \log_{4}2x & = \frac{\log_{2}2x}{\log_{2}4}\\ & =\frac{\log_{2}{2}+\log_{2}{x}}{2}\\ & =\frac{1+u}{2} \\ \end{align}$

12) $\log_{8}\frac{4}{\sqrt{x}}\,$ $=\frac{4+u}{6}$

- $\begin{align} \log_{8}\frac{4}{\sqrt{x}} & = \frac{\log_{2}\frac{4}{\sqrt{x}}}{\log_{2}8}\\ & = \frac{\log_{2}4-\frac{1}{2}\log_{2}x}{3}\\ & =\frac{2+\frac{1}{2}u}{3} \\ & =\frac{4+u}{6} \\ \end{align}$

13) $\log_{4x}2\,$ $=\frac{1}{2+u}$

- $\begin{align} \log_{4x}2 & = \frac{1}{\log_{2}4x}\\ & =\frac{1}{\log_{2}4+\log_{2}x}\\ & =\frac{1}{2+u} \\ \end{align}$

14) $\left(\log_{2}x\right)^{2}$ $=u^{2}\,$

Value of $a^{\log_{a}x}$

h) $a^{\log_{a}x}$ $=x\,$

Prove :
Once again, we can't manipulate by itself, thus first we put something on the right
- $\mbox{Let } a^{\log_{a}x}=y$
- Now, we can change it to log form ${\color{Red}a^{{\color{Blue}\log_{a}x}}}=y$
- $\log_{{\color{Red}a}}y={\color{Blue}\log_{a}x}$ Hmm....
- $\therefore y=x$ But what is $y\,$ ?
- $\therefore a^{\log_{a}x}=x$

i) $e^{\ln x}\,$ $=x\,$

Note : $a^{n\log_{a}x}$ $=a^{\log_{a}x^{n}}$ $=x^{n}\,$

Exercise 4

1) $3^{\log_{3}4}$ $=4\,$

2) $2^{-\log_{2}3}$ $=\frac{1}{3}$

- $2^{-\log_{2}3}=2^{\log_{2}3^{-1}}=\frac{1}{3}$

3) $10^{\lg 4}\,$ $=4\,$

4) $10^{-2\lg x}\,$ $=\frac{1}{x^{2}}$

- $10^{-2\lg x}=10^{\lg x^{-2}}=\frac{1}{x^{2}}$

5) $e^{-3\ln 2}\,$ $=\frac{1}{8}$

- $e^{-3\ln 2}=e^{\ln 2^{-3}}=\frac{1}{8}$

6) $e^{-\frac{1}{2}\ln x}\,$ $=\frac{1}{\sqrt{x}}$

- $e^{-\frac{1}{2}\ln x}=e^{\ln x^{-\frac{1}{2}}}=\frac{1}{\sqrt{x}}$

Same Base

If $\log_{a}x=\log_{a}y\,$ $\therefore x=y$
If $\log_{a}N=x\,$ $\therefore a^{x}=N\,$

Thus, to solve logarithm equations with same/similar base, we rearrange the equation into

single logarithm on both sides OR
single logarithm on one side and number on the other side

Note : Check that

base $>0\,$ and $N>0\,$
answer satisfies equation

Examples

$\log_{2}\left(x-3\right)+\log_{2}5=\log_{2}2x$
- $\log_{2}\left(x-3\right)+\log_{2}5=\log_{2}2x$ Combine the log terms
- $\log_{2}5\left(x-3\right)=\log_{2}2x$ Alright, now we can simplify it
- $\therefore 5\left(x-3\right)=2x$ Now it just becomes a simple equation
- $\begin{align} &5x-15=2x \\ &3x=15 \\ &x=5 \\ \end{align}$
- CHECK

$4\log_{9}x=1+\log_{3}\left(x+6\right)$
- $4\log_{9}x=1+\log_{3}\left(x+6\right)$
- $4\frac{\log_{3}x}{\log_{3}9}=1+\log_{3}\left(x+6\right)$
- $2\log_{3}x-\log_{3}\left(x+6\right)=1$ Combine to a single log carefully...
- $\log_{3}\frac{x^{2}}{\left(x+6\right)}=1$ Change to index form carefully
- $\therefore \frac{x^{2}}{\left(x+6\right)}=3$
- $x^{2}=3x+18\,$
- $x^{2}-3x-18=0\,$
- $x=-3\mbox{(rejected)} \mbox{ or } x=6\,$
- $\therefore x=6$
- CHECK $\begin{align}& 4\log_{9}6=3.26 \\ & 1+\log_{3}12=3.26 \\\end{align}$

Exercise 5

Solve the following equations

1) $\log_{2}x=1-\log_{2}8x\,$ $x=\frac{1}{2}\,$

- $\begin{align} & \log_{2}x=1-\log_{2}8x \\ & \log_{2}x+\log_{2}8x=1 \\ & \log_{2}8x^{2}=1 \\ & \therefore 8x^{2}=2 \\ & x^{2}=\frac{1}{4} \\ & x=\pm \frac{1}{2} \\ & x=\frac{1}{2} \mbox{ or } x=-\frac{1}{2}\mbox{(rejected)}\\ & \therefore x=\frac{1}{2}\\ \end{align}$
- Check : $\log_{2}\frac{1}{2}=-1; 1-\log_{2}4=-1$

2) $\lg\left(x-4\right)+\lg\left(x-5\right)=\lg 6\,$ $x=7\,$

- $\begin{align} & \lg\left(x-4\right)+\lg\left(x-5\right)=\lg 6 \\ & \lg\left(x-4\right)\left(x-5\right)=\lg 6 \\ & \therefore \left(x-4\right)\left(x-5\right) =6 \\ & x^{2}-9x+20=6 \\ & x^{2}-9x+14=0 \\ & \left(x-7\right)\left(x-2\right)=0\\ & x=7 \mbox{ or } x=2\mbox{(rejected)}\\ & \therefore x=7 \end{align}$
- Check : $\lg3+\lg2=0.778; \lg6=0.778\,$

3) $\ln\left(x+2\right)=2+\ln2$ $x=2e^{2}-2\,$

- $\begin{align} & \ln\left(x+2\right)=2+\ln2 \\ & \ln\left(x+2\right)-\ln2=2 \\ & \frac{\left(x+2\right)}{2}=e^{2} \\ & x+2 =2e^{2} \\ & x =2e^{2}-2 \\ \end{align}$
- Check : $\ln \left(2e^{2}\right)=2.693 ; 2+\ln 2=2.693$

4) $\log_{x}16-4\log_{x}3=4\,$ $x=\frac{2}{3}\,$

- $\begin{align} & \log_{x}16-4\log_{x}3=4 \\ & \log_{x}\frac{16}{81}=4 \\ & \frac{16}{81}=x^{4} \\ & x = \sqrt[4]{\frac{16}{81}} \\ &\therefore x=\frac{2}{3} \end{align}$
- Check : $\log_{\frac{2}{3}}16-4\log_{\frac{2}{3}}3=4$

5) $\log_{\sqrt{8}}\left(x-5\right)=\frac{2}{3}+\log_{8}\left(x-6\right)$ $x=7\,$

- $\begin{align} & \log_{\sqrt{8}}\left(x-5\right)=\frac{2}{3}+\log_{8}\left(x-6\right) \\ & \frac{\log_{8}\left(x-5\right)}{\log_{8}{\sqrt{8}}}=\frac{2}{3}+\log_{8}\left(x-6\right) \\ & \frac{\log_{8}\left(x-5\right)}{\frac{1}{2}}-\log_{8}\left(x-6\right)=\frac{2}{3} \\ & 2\log_{8}\left(x-5\right)-\log_{8}\left(x-6\right)=\frac{2}{3} \\ & \log_{8}\frac{\left(x-5\right)^{2}}{\left(x-6\right)}=\frac{2}{3} \\ & \therefore \frac{\left(x-5\right)^{2}}{\left(x-6\right)} = 8^{\frac{2}{3}} \\ & \frac{\left(x-5\right)^{2}}{\left(x-6\right)} = 4 \\ & x^{2}-10x+25=4x-24 \\ & x^{2}-14x+49=0 \\ & \left(x-7\right)^{2}=0 \\ & \therefore x=7 \end{align}$
- Check : $\log_{\sqrt{8}}2=\frac{2}{3};\frac{2}{3}+\log_{8}1=\frac{2}{3}$

6) $\frac{1}{2}\log_{3}\frac{x+3}{x-2}=\log_{9}{\left(x-5\right)}$ $x=7\,$

- $\begin{align} & \frac{1}{2}\log_{3}\frac{x+3}{x-2}=\log_{9}{\left(x-5\right)}\\ & \frac{1}{2}\log_{3}\frac{x+3}{x-2}=\frac{\log_{3}{\left(x-5\right)}}{\log_{3}9}\\ & \frac{1}{2}\log_{3}\frac{x+3}{x-2}=\frac{1}{2}\log_{3}{\left(x-5\right)} \\ & \therefore \frac{x+3}{x-2} = x-5 \\ & x+3 = \left(x-2\right)\left(x-5\right) \\ & x+3 =x^{2}-7x+10 \\ & x^{2}-8x+7=0 \\ & \left(x-7\right)\left(x-1\right)=0\\ & x=7 \mbox{ or } x=1\mbox{(rejected)}\\ & \therefore x=7 \end{align}$
- Check : $\frac{1}{2}\log_{3}2=0.315;\log_{9}{2}=0.315$

7) $\log_{6}\frac{x^{2}-5x}{x-9}=0$ $x=3\,$

- $\begin{align} & \log_{6}\frac{x^{2}-5x}{x-9}=0 \\ & \therefore \frac{x^{2}-5x}{x-9} = 1 \\ & x^{2}-5x = x-9 \\ & x^{2}-6x+9 = 0 \\ & \left(x-3\right)^{2}=0\\ & \therefore x=3 \end{align}$
- Check : $\log_{6}\frac{-6}{-6}=0$

Using Substitution

When necessary, we use substitution. For example, when we have both $\log_{x}3\,$ & $\log_{3}x\,$ , we know we can change one to the other easily.

Note that the base and the number must be positive, but the index can be positive or negative.

Examples

$3\log_{8}x=\log_{x}64+5\,$
Let's study what we have here
- $3{\color{Blue}\log_{8}x}={\color{Red}\log_{x}64}+5\,$
- $3\log_{8}x=\log_{x}64+5\,$
- $\mbox{Let } u=\log_{8}x\,$
- $3u=\frac{2}{u}+5$
- $3u^{2}=2+5u\,$
- $3u^{2}-5u-2=0\,$
- $\left(3u+1\right)\left(u-2\right)=0$
- $u=-\frac{1}{3} \mbox{ or } u=2$
- $x=8^{-\frac{1}{3}} \mbox{ or } x=8^{2}$
- $x=\frac{1}{2} \mbox{ or } x=64$
- CHECK: $\begin{align}& 3\log_{8}\frac{1}{2}=-1\\ & \log_{\frac{1}{2}}64+5=-1\\\end{align}$ $\begin{align}& 3\log_{8}64=6\\ & \log_{64}64+5=6\\\end{align}$

$\log_{3}x=\log_{x}3\,$
- $\log_{3}x=\frac{1}{\log_{3}x}$ If needed, we can use substitution
- $\log_{3}x=\pm 1$
- $\log_{3}x=1 \mbox { or } \log_{3}x=-1\,$
- $x=3 \mbox { or } x=\frac{1}{3}$
- CHECK: $\log_{3}3=\log_{3}3\,$ , $\log_{3}\frac{1}{3}=-1,\log_{\frac{1}{3}}3=-1$

Exercise 6

Solve the following equations

1) $\log_{2}x=\log_{x}16\,$ $x=4 \mbox{ or } x=\frac{1}{4}$

- $\begin{align} & \log_{2}x=\log_{x}16 \\ & \log_{2}x=4\log_{x}2 \\ & \log_{2}x=\frac{4}{\log_{2}x} \\ & \left(\log_{2}x\right)^{2}=4 \\ & \log_{2}x=\pm 2 \\ & \log_{2}x= 2 \mbox{ or } \log_{2}x= -2 \\ & x=4 \mbox{ or } x=\frac{1}{4} \end{align}$
- Check : $\log_{2}4=2,\log_{4}16=2 ; \log_{2}\frac{1}{4}=-2,\log_{\frac{1}{4}}16=-2$

2) $\log_{x}5+1=6\log_{5}x\,$ $x=\frac{1}{\sqrt[3]{5}} \mbox{ or } x=\sqrt{5}$

- $\begin{align} & \log_{x}5+1=6\log_{5}x \\ & \frac{1}{\log_{5}x}+1=6\log_{x}5 \\ & \mbox { Let } u=\log_{5}x \\ & \frac{1}{u} +1 =6u \\ & 1+u=6u^{2} \\ & 6u^{2}-u-1=0\\ & \left(3u+1\right)\left(2u-1\right)=0 \\ & u= -\frac{1}{3} \mbox{ or } u= \frac{1}{2} \\ & \log_{5}x= -\frac{1}{3} \mbox{ or } \log_{5}x= \frac{1}{2} \\ & x=\frac{1}{\sqrt[3]{5}} \mbox{ or } x=\sqrt{5}\\ \end{align}$
- Check : $\log_{\frac{1}{\sqrt[3]{5}}}5+1=-2, 6\log_{5}{\frac{1}{\sqrt[3]{5}}}=-2; \log_{\sqrt{5}}5+1=\frac{3}{2}, 6\log_{5}{\sqrt{5}}=-2$

3) $2\left(3\log_{x}4-\log_{4}x\right)=1\,$ $x=8 \mbox{ or } x=\frac{1}{16}$

- $\begin{align} & 2\left(3\log_{x}4-\log_{4}x\right)=1\\ & 2\left(\frac{3}{\log_{4}x}-\log_{4}x\right)=1\\ & \mbox { Let } u=\log_{4}x \\ & 2\left(\frac{3}{u}-u\right)=1 \\ & \frac{6}{u}-2u=1 \\ & 6-2u^{2}=u \\ & 2u^{2}+u-6 = 0 \\ & \left(2u-3\right)\left(u+2\right)=0 \\ & u= \frac{3}{2} \mbox{ or } u= -2 \\ & \log_{4}x= \frac{3}{2} \mbox{ or } \log_{4}x= -2 \\ & \therefore x=4^{\frac{3}{2}} \mbox{ or } x=4^{-2}\\ & x=8 \mbox{ or } x=\frac{1}{16}\\ \end{align}$
- Check : $2\left(3\log_{8}4-\log_{4}8\right)=1; 2\left(3\log_{\frac{1}{16}}4-\log_{4}\frac{1}{16}\right)=1$

4) $\log_{x}3+\log_{9}x=\frac{3}{2}\,$ $x=9 \mbox{ or } x= 3\,$

- $\begin{align} & \log_{x}3+\log_{9}x=\frac{3}{2} \\ & \log_{x}3+\frac{\log_{3}x}{\log_{3}9}=\frac{3}{2} \\ & \log_{x}3+\frac{\log_{3}x}{2}=\frac{3}{2} \\ & 2\log_{x}3+\log_{3}x=3 \\ & \frac{2}{\log_{3}x}+\log_{3}x=3 \\ & \mbox { Let } u=\log_{3}x \\ & \frac{2}{u} +u =3 \\ & 2+u^{2}=3u \\ & u^{2}-3u+2=0\\ & \left(u-2\right)\left(u-1\right)=0 \\ & u= 2 \mbox{ or } u= 1 \\ & \log_{3}x= 2 \mbox{ or } \log_{3}x= 1 \\ & x=9 \mbox{ or } x=3 \\ \end{align}$
- Check : $\log_{9}3+\log_{9}9=\frac{3}{2};\log_{3}3+\log_{9}3=\frac{3}{2}$

5) $\log_{2}4x=\log_{x}8\,$ $x=\frac{1}{8} \mbox{ or } x=2 \,$

- $\begin{align} & \log_{2}4x=\log_{x}8 \\ & \log_{2}4+\log_{2}x=3\log_{x}2 \\ & 2+\log_{2}x=\frac{3}{\log_{2}x} \\ & \mbox { Let } u=\log_{2}x \\ & 2+u=\frac{3}{u} \\ & 2u+u^{2}=3 \\ & u^{2}+2u-3=0\\ & \left(u+3\right)\left(u-1\right)=0 \\ & u= -3 \mbox{ or } u= 1 \\ & \log_{2}x= -3 \mbox{ or } \log_{2}x= 1 \\ & x=\frac{1}{8} \mbox{ or } x=2 \\ \end{align}$
- Check : $\log_{2}{\frac{1}{2}}=-1,\log_{\frac{1}{8}}8=-1;\log_{2}8=\log_{2}8$

Using Logarithm to Solve Exponent Equations

When the base is different, we have no choice but to use logarithm

Writing the final answer

How we write the final answer depends on the question. A careless mistake can result in loss of marks.

Precise - $\ln 2, e^{2}, 2\lg 3\,$
3 significant figures - ${\color{Blue}1}.{\color{Blue}24}, 0.{\color{Blue}354}, {\color{Blue}12}.{\color{Blue}1}$
3 decimal places - $1.{\color{Blue}243}, 0.{\color{Blue}354}, 12.{\color{Blue}143}$

Note that we do need to round off

$1.243 \to$ $1.24\,$ (3sf)
$1.247 \to$ $1.25\,$ (3sf)

Example

Solve for $x\,$ (3sf)

$5^{x}=2\,$
Analyze :
- $5^{x}=2\,$
- Normally we just use log base 10
- $\lg 5^{x}=\lg2\,$ We need the $\lg 5^{{\color{Red}x}}=\lg2\,$
- $x\lg 5=\lg 2\,$
- $x=\frac{\lg 2}{\lg 5}$
- $x=0.431 \mbox{(3 s.f.)}\,$
- Check :

$2^{x}\cdot 3^{x}=7\,$
Analyze :
- $2^{x}\cdot 3^{x}=7\,$ We know that $a^{n}b^{n}=\,$ $\left(ab\right)^{n}$
- $6^{x}=7\,$
- $\lg 6^{x}=\lg 7\,$
- $x\lg 6=\lg 7\,$
- $x=\frac{\lg 7}{\lg 6}\,$
- $x=1.09 \mbox{(3 s.f.)}\,$
- Check :
  $x={\color{Red}1}.{\color{Red}09}$
  $2^{1.09}\cdot 3^{1.09}=7.05$

$3^{x-2}=5\,$
Analyze :
- $3^{{\color{Red}x}-2}=5\,$ We just need the $x\,$
- $\lg 3^{x}=\lg 45\,$
- $x\lg 3=\lg 45\,$
- $x=\frac{\lg 45}{\lg 3}$
- $x=3.46 \mbox{(3 s.f.)}\,$
- Check : $3^{1.46}=4.97\,$

$2^{3x+1}=3\,$
- $2^{3x+1}=3\,$
- $2\left(2^{3x}\right)=3$
- $2^{3x}=\frac{3}{2}$
- $x\lg 8=\lg \frac{3}{2}$
- $x=\frac{\lg \frac{3}{2}}{\lg 8}$
- $x=0.195 \mbox{(3 s.f.)}\,$
- Check : $2^{3\left(0.195\right)+1}=3\,$

More Rearranging

$2^{2x}\cdot 7^{x}=$ $4^{x} \cdot 7^{x}=28^{x}$
$\frac{3^{x-2}}{2^{3x}}=$ $\frac{3^{x}}{4\left(8^{x}\right)}=\frac{1}{4}\left(\frac{3}{8}\right)^{x}$
$\frac{2^{x-3}\cdot 3^{2x+1}}{5^{x-1}}=$ $\frac{\frac{2^{x}}{8}\cdot 3\left(3^{2x}\right)}{\frac{5^{x}}{5}}=\frac{15}{8}\left(\frac{18}{5}\right)^{x}$

Example

$4^{x}\cdot 5^{x-3}=3^{x-2}$ (4 s.f.)
Trying to use the SPM method will be suicide...
- $4^{x}\cdot 5^{x-3}=3^{x-2}$
- $4^{x}\cdot \frac{5^{x}}{125}=\frac{3^{x}}{9}$
- $\frac{4^{x}\cdot 5^{x}}{3^{x}}=\frac{125}{9}$
- $\left(\frac{20}{3}\right)^{x}=\frac{125}{9}$
- $x\lg\left(\frac{20}{3}\right)=\lg \left(\frac{125}{9}\right)$
- $x=\frac{\lg \left(\frac{125}{9}\right)}{\lg\left(\frac{20}{3}\right)}$
- $x=1.387 \mbox{(4 s.f.)}\,$
- Check : $4^{1.387}\cdot 5^{-1.613}=0.51; 3^{-0.613}=0.51$

Identifying Type of Exponent Equations

As always, the difficulty isn't with the solving, it is deciding which method to use for a particular question.

Arrange to
- Note the equal base : ${\color{Blue}3}^{2x+1}-{\color{Blue}3}^{x-2}=0$
- The two terms : ${\color{Red}3^{2x+1}}-{\color{Red}3^{x-2}}=0$ (the zero doesn't matter)
- Arrange to ${\color{Blue}3}^{m}={\color{Blue}3}^{n}\,$ , not that the important thing is to simplify both sides to a single exponent, the index can be as complicated as needed

Substitute
- Note that it is not only two terms, so we can't do like above even though it is equal base
- ${\color{Blue}3^{2x+1}}+{\color{Red}3^{x-2}}+1=0\,$ is different, but we know that both can be changed to in terms of $3^{x}\,$
- Thus we will "break apart" the index such that we get only $u={\color{Red}3^{x}}\,$

Use log, arrange to
- Note the different base
- Thus we have to use log, and rearrange to $\heartsuit^{{\color{Red}x}}=\bigcirc$
- Note that the only important thing is to get the $x\,$ , the two numbers can be as complicated as needed

Use log, arrange to
- The single $2\,$ will make this different base

Exercise 7

Solve the following equations, leaving your answers correct to 3 s.f.

1) $4^{3x-2}=15\,$ $x=1.32\,$

- $\begin{align} & 4^{3x-2}=15 \\ & \frac{4^{3x}}{16}=15\\ & 64^{x}=240 \\ & \therefore \lg 64^{x} =\lg 240 \\ & x\lg 64 =\lg 240 \\ & x=\frac{\lg 240}{\lg 64} \\ & x= 1.32 \mbox { (3sf)} \end{align}$
- Check : $4^{1.96}=15.13\,$

2) $3^{x}\cdot2^{x+1}=5\,$ $x=0.511\,$

- $\begin{align} & 3^{x}\cdot2^{x+1}=5 \\ & 3^{x}\cdot2\cdot2^{x}=5\\ & 6^{x}=\frac{5}{2} \\ & \therefore \lg 6^{x} =\lg \frac{5}{2} \\ & x\lg 6 =\lg \frac{5}{2} \\ & x=\frac{\lg \frac{5}{2}}{\lg 6} \\ & x= 0.511 \mbox { (3sf)} \end{align}$
- Check : $3^{0.511}\cdot2^{1.511}=4.996$

3) $2^{x}\cdot5^{x}=3^{x+2}\,$ $x=1.82\,$

- $\begin{align} & 2^{x}\cdot5^{x}=3^{x+2} \\ & 2^{x}\cdot5^{x}=9\cdot3^{x}\\ & \left(\frac{10}{3}\right)^{x}=9 \\ & \therefore \lg \left(\frac{10}{3}\right)^{x}=\lg 9 \\ & x\lg\frac{10}{3} =\lg 9 \\ & x=\frac{\lg 9}{\lg \frac{10}{3}} \\ & x= 1.82 \mbox { (3sf)} \end{align}$
- Check : $2^{1.82}\cdot5^{1.82}=66.07,3^{3.82}=66.5$

4) $3^{x}\cdot4^{1-x}=5^{2x-1}\,$ $x=0.854\,$

- $\begin{align} & 3^{x}\cdot4^{1-x}=5^{2x-1} \\ & 3^{x}\left(\frac{4}{4^{x}}\right)=\frac{5^{2x}}{5}\\ & \frac{3^{x}}{4^{x}\cdot5^{2x}}=\frac{1}{20}\\ & \left(\frac{3}{100}\right)^{x}=\frac{1}{20}\\ & \therefore \lg \left(\frac{3}{100}\right)^{x}=\frac{1}{20} \\ & x\lg\frac{3}{100} =\lg\frac{1}{20} \\ & x=\frac{\lg\frac{1}{20}}{\lg\frac{3}{100}} \\ & x= 0.854 \mbox { (3sf)} \end{align}$
- Check : $3^{0.854}\cdot5^{0.146}=3.13,5^{0.708}=3.123$

5) $2^{x+1}\cdot3^{x-2}=4^{x-2}\,$ $x=-3.13\,$

- $\begin{align} & 2^{x+1}\cdot3^{x-2}=4^{x-2} \\ & 2\cdot2^{x}\left(\frac{3^{x}}{9}\right)=\frac{4^{x}}{16}\\ & \frac{2^{x}\cdot3^{x}}{4^{x}}=\frac{9}{32}\\ & \left(\frac{3}{2}\right)^{x}=\frac{9}{32}\\ & \therefore \lg \left(\frac{3}{2}\right)^{x}=\frac{9}{32} \\ & x\lg\frac{3}{2} =\lg\frac{9}{32} \\ & x=\frac{\lg\frac{9}{32}}{\lg\frac{3}{2}} \\ & x= -3.13 \mbox{ (3sf)} \end{align}$
- Check : $2^{-2.13}\cdot3^{-5.13}=8.15 \times 10^{-4};4^{-5.13}=8.15 \times 10^{-4}$

6) $4^{3x}\cdot5^{2x-1}=6^{x+2}\,$ $x=0.930\,$

- $\begin{align} & 4^{3x}\cdot5^{2x-1}=6^{x+2} \\ & 4^{3x}\cdot\left(\frac{5^{2x}}{5}\right)=36\cdot6^{x}\\ & \frac{4^{3x}\cdot5^{2x}}{6^{x}}=180\\ & \left(\frac{1600}{6}\right)^{x}=180\\ & \therefore \lg \left(\frac{1600}{6}\right)^{x}=180 \\ & x\lg\frac{1600}{6} =\lg 180 \\ & x=\frac{\lg 180}{\lg\frac{1600}{6}} \\ & x= 0.930 \mbox{ (3sf)} \end{align}$
- Check : $4^{2.79}\cdot5^{0.86}=190.9; 6^{.930}=190.5$

7) $2\left(9^{x}\right)-7\left(3^{x}\right)+3=0$ $x= -0.631 \mbox{ or } x=1\,$

- $\begin{align} & 2\left(9^{x}\right)-7\left(3^{x}\right)+3=0 \\ & 2\left(3^{x}\right)^{2}-7\left(3^{x}\right)+3=0 \\ & \mbox {Let } u=3^{x} \\ & 2u^{2}-7u+3=0 \\ &\left(2u-1\right)\left(u-3\right)=0 \\ & u=\frac{1}{2} \mbox{ or } u=3 \\ & \therefore 3^{x}=\frac{1}{2} \mbox{ or } 3^{x}=3 \\ & x=\frac{\lg\frac{1}{2}}{\lg 3} \mbox{ or } x=1 \\ & x= -0.631 \mbox{ (3sf)} \mbox{ or } x=1 \end{align}$
- Check : $\begin{align} & 2\left(9^{-0.631}\right)-7\left(3^{-0.631}\right)+3=0\\ & 2\left(9\right)-3\left(7\right)+3=0 \end{align}$

8) $2^{x}+1=12\left(2^{-x}\right)$ $x=1.58\,$

- $\begin{align} & 2^{x}+1=12\left(2^{-x}\right) \\ & 2^{x}+1=\frac{12}{2^{x}} \\ & \mbox {Let } u=2^{x} \\ & u+1=\frac{12}{u} \\ & u^{2}+u=12 \\ & u^{2}+u-12=0 \\ &\left(u+4\right)\left(u-3\right)=0 \\ & u=-4 \mbox{ or } u=3 \\ & \therefore 2^{x}=-4\mbox{ (rejected) } \mbox{ or } 2^{x}=3 \\ & x=\frac{\lg 3}{\lg 2} \\ & x= 1.58 \mbox{ (3sf)} \end{align}$
- Check : $2^{1.58}+1=3.99; 12\left(2^{-1.58}\right)=4.01$

Solving Simultaneous Equations

In simultaneous equation, the main step is always

If possible (and depending whether it is simpler to do substitution), simplify and rearrange one of the equations such that one variable is in terms of the other, then substitute it to the other equation
- $3^{x-3}=3^{y-2}\,$ $\therefore x-3=y-2$ and then rearrange
- $\log_{2}\left(x-y\right)=4\,$ $\therefore x-y=16$ and then rearrange
- $\log_{x}y=4\,$ $\therefore y=x^{4}$

Use substitution when necessary (look for the same/similar variable)
- - ${\color{Red}2}^{{\color{Red}x}+1}=11-{\color{Blue}3}^{{\color{Blue}y}-1} \mbox{ and } {\color{Red}2}^{{\color{Red}x}-2}+{\color{Blue}3}^{{\color{Blue}y}+1}=82$
    Use substitution $a={\color{Red}2^{x}}, b={\color{Blue}3^{x}}$
- - ${\color{Red}\log_{2}x}+{\color{Blue}\log_{2}y}=2 \mbox{ and } {\color{Red}\log_{2}x}=3+{\color{Blue}\log_{2}y}$
    Use substitution $a={\color{Red}\log_{2}x}, b={\color{Blue}\log_{2}y}$
    Note also that here, even though it is possible to rearrange one equation to $x\,$ in terms of $y\,$ and substitute to the other, substitution is much easier
- - Note that $\log_{4}xy^{2}=-1\,$ can be changed to in terms of $\log_{2}x\,$ and $\log_{2}y\,$ , which also exists in the second equation
    Thus, we will use substitution $a={\color{Red}\log_{2}x}, b={\color{Blue}\log_{2}y}$ , after rearranging the first equation into terms of $\log_{2}x\,$ and $\log_{2}y\,$
    Note that we WON'T rearrange the first one to $xy^{2}=\frac{1}{4}$ as substitution will be much easier in this case

Examples

$3^{2-y}=3^{x-1} \mbox { and } \log_{2}4x+\log_{2}y=3\,$
Analyze :
- $\begin{align} & 3^{2-y}=3^{x-1} \\ & \therefore 2-y=x-1 \\ & x=3-y \\ \end{align}$
- $\begin{align} & \log_{2}4\left(3-y\right)+\log_{2}y=3 \\ & \log_{2}4y\left(3-y\right)=3 \\ & \therefore 4y\left(3-y\right)= 8 \\ & 12y-4y^{2}=8 \\ & y^{2}-3y+2=0 \\ & \left(y-2\right)\left(y-1\right)\\ & \therefore y=2, y=1 \\ \end{align}$
- $\mbox{When } y=2, x=1, \mbox{When } y=1, x=2 \,$
- $\therefore x=1, y=2 \mbox{ or } x=2, y=1 \,$

$\log_{5}xy=2 \mbox { and } 3\log_{5}x-\log_{5}y=-6\,$
Analyze :
- $\begin{align} & \log_{5}xy=2 \\ & \log_{5}x+\log_{5}y=2 \\ \end{align}$
- Thus, we now see $\begin{align} & \log_{5}xy=2 & 3{\color{Red}\log_{5}x}-{\color{Blue}\log_{5}y}=-6 \\ & {\color{Red}\log_{5}x}+{\color{Blue}\log_{5}y}=2 \\ \end{align}$
- $\mbox{Let } a=\log_{5}x, b=\log_{5}y \,$
- $\therefore a+b=2 \qquad 3a-b=-6$
- We then solve for $a\,$ and $b\,$
- $\begin{align} & a+b=2\frac{\quad}{}(1) \qquad 3a-b=-6\frac{\quad}{}(2) \\ & (1)+(2): \\ & 4a=-4, \therefore a =-1, b=3 \\ \end{align}$
- And thus, solve for $x\,$ and $y\,$
- $\therefore \log_{5}x=-1, \log_{5}y=3$
- $\therefore x=\frac{1}{5}, y=125$
- $\log_{5}25=2 \mbox { and } 3\log_{5}\frac{1}{5}-\log_{5}125=-3-3=-6\,$

$\left(\log_{8}x\right)\left(\log_{8}y\right)=-2 \mbox { and } \log_{8}xy^{3}=1\,$
Analyze :
- $\begin{align} \left(\log_{8}x\right)\left(\log_{8}y\right)=-2 \qquad\qquad \log_{8}xy^{3}=1\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \log_{8}x+3\log_{8}y=1\\ \end{align}$
- $\mbox{Let } a=\log_{8}x, b=\log_{8}y\,$
- $\begin{align} & ab=-2\frac{\quad}{}(1) \qquad a+3b=1 \\ & \qquad \qquad\qquad \qquad a=1-3b\frac{\quad}{}(2) \\ & (2) \to (1):\\ & \left(1-3b\right)b=-2 \\ & b-3b^{2}=-2 \\ & 3b^{2}-b-2=0 \\ & \left(3b+2\right)\left(b-1\right)\\ & \therefore b=-\frac{2}{3} \mbox{ or } b=1 \\ & \mbox{When } b=-\frac{2}{3}, a =3 \\ & \mbox{When } b=1, a =-2 \\ \end{align}$
- $\begin{align} \therefore \log_{8}x =3,&\quad \log_{8}y =-\frac{2}{3} \\ x=512, &\quad y=\frac{1}{4} \end{align}$
- $\begin{align} \mbox{Or } \log_{8}x =1&\quad \log_{8}y =-2 \\ x=8, &\quad y=\frac{1}{64} \end{align}$
- $x=512,y=\frac{1}{4} \mbox{ or } x=8,y=\frac{1}{64}$
- Check :
  $x=512,y=\frac{1}{4}; \left(\log_{8}512\right)\left(\log_{8}\frac{1}{4}\right)=3\times-\frac{2}{3}=-2 \mbox { and } \log_{8}8=1$
  $x=8,y=\frac{1}{64}; \left(\log_{8}8\right)\left(\log_{8}\frac{1}{64}\right)=1\times-2=-2 \mbox { and } \log_{8}8=1$

$\log_{3}\left(x+y\right)=2 \mbox { and } \log_{3}x+\log_{3}y=2+\log_{3}2\,$
Analysis :
- $\begin{align} & \log_{3}\left(x+y\right)=2 \\ & \therefore x+y=9 \\ & y=9-x \end{align}$
- $\begin{align} & \log_{3}\left(x+y\right)=2 \qquad \log_{3}x+\log_{3}y=2+\log_{3}2\frac{\quad}{}(2)\\ & \therefore x+y=9 \\ & y=9-x\frac{\quad}{}(1)\\ & (1)\to(2):\\ & \log_{3}x+\log_{3}\left(9-x\right)=2+\log_{3}2 \\ & \log_{3}\left[\frac{x\left(9-x\right)}{2}\right]=2 \\ & \frac{x\left(9-x\right)}{2} =9 \\ & 9x-x^{2}=18 \\ & x^{2}-9x+18=0 \\ & \left(x-6\right)\left(x-3\right)=0 \\ & x=6 \mbox{ or } x=3 \\ & \mbox{When } x=6, y=3\\ & \mbox{When } x=3, y=6 \\ & \therefore x=6, y=3 \mbox{ or } x=3, y=6 \\ \end{align}$
- Check :
  $x=6, y=3 ;\; \log_{3}9=2 \mbox { and } \log_{3}6+\log_{3}3=2.631, 2+\log_{3}2=2.631\,$
  $x=3, y=6 ;\; \log_{3}9=2 \mbox { and } \log_{3}3+\log_{3}6=2.631 \,$

Exercise 8

1) $3^{x}=9\cdot3^{3-y} \mbox{ and }\log_{4}x+\log_{4}y=1$ $y=4, x=1 \mbox{ or } y=1, x=4 \,$

- $\begin{array}{ll} 3^{x}=3^{5-y} \qquad & \log_{4}x+\log_{4}y=1\\ x=5-y\frac{\qquad}{}(1) & \log_{4}xy=1\\ & xy=4\frac{\qquad}{}(2) \\ \end{array}$
- $\begin{align} & (1)\to(2):\\ & \left(5-y\right)y=4\\ & 5y-y^{2}=4 \\ & y^{2}-5y+4=0\\ & \left(y-4\right)\left(y-1\right)\\ & y=4 \mbox{ or } y=1 \\ & \mbox{When } y=4, x=1 \\ & \mbox{When } y=1, x=4 \\ & \therefore y=4, x=1 \mbox{ or } y=1, x=4 \\ \end{align}$
- Check : $3=9\left(\frac{1}{3}\right), \log_{4}1+\log_{4}4=0+1=1; 3^{4}=9\left(9\right) \mbox{ and }\log_{4}4+\log_{4}1=1+0=0$

2) $xy=4 \mbox{ and } \log_{2}y-4\log_{2}x=2\,$ $x=1, y=4\,$

- $\begin{align} & xy=4 \qquad \log_{2}y-4\log_{2}x=2\frac{\qquad}{}(2)\\ & y=\frac{4}{x}\frac{\qquad}{}(1) & \\ & (1)\to(2):\\ & \log_{2}\left(\frac{4}{x}\right)-4\log_{2}x=2 \\ & \log_{2}\frac{\left(\frac{4}{x}\right)}{x^{4}}=2\\ & \log_{2}\frac{4}{{x}^{5}}=2\\ & \therefore \frac{4}{{x}^{5}}=4 \\ & {x}^{5}=1\\ & x=1, \therefore y=4 \end{align}$ or $\begin{align} & xy=4 \qquad \qquad \quad \log_{2}y-4\log_{2}x=2\\ & y=\frac{4}{x}\frac{\qquad}{}(1) \qquad \log_{2}\left(\frac{y}{x^{4}}\right)=2\\ & \qquad \qquad \qquad\qquad \frac{y}{x^{4}}=4\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & \frac{\left(\frac{4}{x}\right)}{x^{4}}=4\\ & \frac{4}{{x}^{5}}=4 \\ & {x}^{5}=1\\ & x=1, \therefore y=4 \end{align}$
- Check : $1\left(4\right)=4 \mbox{ and } \log_{2}4-4\log_{2}1=2-4\left(0\right)=2$

3) $\log_{2}x+\log_{2}y=8 \mbox{ and } \log_{2}x=3\log_{2}y\,$ $x=64 , y=4\,$

- $\begin{align} & \log_{2}x+\log_{2}y=8 \mbox{ and } \log_{2}x=3\log_{2}y\ \\ & \mbox{Let } a =\log_{2}x, b=\log_{2}y \\ & \therefore a+b=8\frac{\qquad}{}(1) , a=3b\frac{\qquad}{}(2)\\ & (2)\to(1):\\ & 4b=8, \therefore b=2, \therefore, a=6 \\ & \therefore \log_{2}x=6, \log_{2}y=2 \\ & \therefore x=64 , y=4 \\ \end{align}$
- Check : $\log_{2}64+\log_{2}4=8 \mbox{ and } \log_{2}64=3\log_{2}4\,$

4) $\log_{2}x+\log_{2}y=3 \mbox{ and } \log_{2}\left(\frac{x}{y^{2}}\right)=9\,$ $x=32 , y=\frac{1}{4}$

- $\begin{align} & \log_{2}x+\log_{2}y=3 \mbox{ and } \log_{2}\left(\frac{x}{y^{2}}\right)=9\ \\ & \qquad \qquad \qquad \qquad \qquad \log_{2}x-2\log_{2}y=9 \\ & \mbox{Let } a =\log_{2}x, b=\log_{2}y \\ & \therefore a+b=3\frac{\qquad}{}(1) , a-2b=9\frac{\qquad}{}(2)\\ & (2)-(1):\\ & 3b=-6, \therefore b=-2, \therefore a=5 \\ & \therefore \log_{2}x=5, \log_{2}y=-2 \\ & \therefore x=32 , y=\frac{1}{4} \\ \end{align}$
- Check : $\log_{2}32+\log_{2}\frac{1}{4}=5-2=3 \mbox{ and } \log_{2}\left(\frac{32}{\left(\frac{1}{4}\right)^{2}}\right)=9$

5) $\log_{4}xy=1 \mbox{ and } \left(\log_{2}x\right)\left(\log_{2}y\right)=-3\,$ $x=8, y=\frac{1}{2} \mbox { or } x=\frac{1}{2}, y=8$

- $\begin{align} & \log_{4}xy=1 \mbox{ and } \left(\log_{2}x\right)\left(\log_{2}y\right)=-3 \\ & \frac{\log_{2}xy}{\log_{2}4 }=1 \\ & \log_{2}xy =2 \\ & \log_{2}x+\log_{2}x =2 \\ & \mbox{Let } a =\log_{2}x, b=\log_{2}y \\ & \therefore a+b=2,\qquad\qquad ab=-3\frac{\qquad}{}(2)\\ & \therefore b=2-a\frac{\qquad}{}(1)\\ & (1)\to(2):\\ & a\left(2-a\right)=-3 \\ & 2a-a^{2}=-3 \\ & a^{2}-2a-3=0 \\ & \left(a-3\right)\left(a+1\right)=0\\ & \therefore a=3 \mbox{ or }a=-1 \\ & \mbox{When } a=3, b=-1 \\ & \mbox{When } a=-1, b=3 \\ & \therefore \log_{2}x=3, \log_{2}y=-1 \mbox { or } \log_{2}x=-1, \log_{2}y=3 \\ & \therefore x=8, y=\frac{1}{2} \mbox { or } x=\frac{1}{2}, y=8\\ \end{align}$
- Check : $\log_{4}4=1 \mbox{ and } \left(\log_{2}8\right)\left(\log_{2}\frac{1}{2}\right)=3\times-1=-3$

6) $\log_{3}xy^{2}=3 \mbox{ and } \left(\log_{3}x\right)\left(\log_{3}y\right)=1\,$ $x=9, y=\sqrt{3} \mbox { or } x=3, y=3$

- $\begin{align} & \log_{3}xy^{2}=3 \mbox{ and } \left(\log_{3}x\right)\left(\log_{3}y\right)=1 \\ & \log_{3}x+2\log_{3}y=3 \\ & \mbox{Let } a =\log_{3}x, b=\log_{3}y \\ & \therefore a+2b=3\frac{\qquad}{}(1)\qquad ab=1\\ & \qquad\qquad\qquad\qquad\qquad \therefore b=\frac{1}{a}\frac{\qquad}{}(2)\\ & (2)\to(1):\\ & a+\frac{2}{a}=3 \\ & a^{2}-3a+2=0 \\ & \left(a-2\right)\left(a-1\right)=0\\ & \therefore a=2 \mbox{ or }a=1 \\ & \mbox{When } a=2, b=\frac{1}{2} \\ & \mbox{When } a=1, b=1 \\ & \therefore \log_{3}x=2, \log_{3}y=\frac{1}{2} \mbox { or } \log_{3}x=1, \log_{3}y=1 \\ & \therefore x=9, y=\sqrt{3} \mbox { or } x=3, y=3\\ \end{align}$
- Check : $\log_{3}27=3 \mbox{ and } \left(\log_{3}9\right)\left(\log_{3}\sqrt{3}\right)=2\times\frac{1}{2}=1$
- $\log_{3}27=3 \mbox{ and } \left(\log_{3}3\right)\left(\log_{3}3\right)=1\,$

7) $\log_{2}x+\log_{2}y=5 \mbox{ and } \log_{y}{x}=\frac{3}{2}\,$ $x=4 , y=8\,$

- $\begin{align} & \log_{2}x+\log_{2}y=5 \mbox{ and } \log{y}{x}=\frac{3}{2}\ \\ & \qquad \qquad \qquad \qquad \qquad \;\frac{\log_{2}x}{\log_{2}y}=\frac{3}{2} \\ & \mbox{Let } a =\log_{2}x, b=\log_{2}y \\ & \therefore a+b=5\frac{\qquad}{}(1)\qquad \frac{a}{b}=\frac{3}{2}\\ & \qquad\qquad\qquad\qquad\qquad\quad a=\frac{3}{2}b\frac{\qquad}{}(2)\\ & (2)\to(1):\\ & \frac{3}{2}b+b=5, \frac{5}{2}b=5,\therefore b=2, \therefore a=3 \\ & \therefore \log_{2}x=2, \log_{2}y=3 \\ & \therefore x=4 , y=8 \\ \end{align}$
- Check : $\log_{2}4+\log_{2}8=2+3=5 \mbox{ and } \log{4}{8}=\frac{3}{2}$

8) $2^{x+2}+2^{y+1}=17 \mbox{ and } 5^{x-1}-5^{y+2}=0\,$ $x=2 , y=-1\,$

- $\begin{align} & 2^{x+2}+2^{y+1}=17\frac{\qquad}{}(1) \qquad 5^{x-1}-5^{y+2}=0 \\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 5^{x-1}=5^{y+2}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \therefore x-1=y+2\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad x=y+3\frac{\qquad}{}(2)\\ & (2)\to(1):\\ & 2^{y+5}+2^{y+1}=17 \\ & 32\left(2^{y}\right)+2\left(2^{y}\right)=17\\ & 34\left(2^{y}\right)=17 \\ & 2^{y}=\frac{1}{2}\\ & \therefore y =-1, x= 2 \end{align}$
- Check : $2^{4}+2^{0}=16+1=17 \mbox{ and } 5^{1}-5^{1}=0\,$

Simple Inequalities

If $a>b\,$

$c>0\,$ (positive), $\therefore ca$ $>\,$ $cb\,$
(positive), $<\,$
- CHANGE SIGN when $\times$ / $\div$ with negative
- DO NOT change when $+\,$ / $-\,$

- $\therefore c^{a}$ $>\,$ $c^{b}\,$ and $\log_{c}a\,$ $>\,$ $\log_{c}b\,$

- $\therefore c^{a}$ $<\,$ $c^{b}\,$ and $\log_{c}a\,$ $<\,$ $\log_{c}b\,$

IMPORTANT

If $a>0\,$

if $x>1\therefore \log_{a}x$ $>0\,$
if $x=1\therefore \log_{a}x$ $=0\,$
if $x<1\therefore \log_{a}x$ $<0\,$

Exercise 9

Complete the following table

1	$-2x>2\,$	$x <-1\,$
2	$2x>-2\,$	$x>-1\,$
3	$-2x>-2\,$	$x<1\,$
4	$x-2>-2\,$	$x>0\,$
5	$x\lg 2>\lg 3\,$	$x>\frac{\lg 3}{\lg 2}$
6	$x\ln \frac{1}{2}>\ln 3\,$	$x<\frac{\ln 3}{\ln \frac{1}{2}}$
7	$x\ln 0.3< \ln 0.1\,$	$x>\frac{\ln 0.1}{\ln 0.3}$
8	$x\lg \frac{5}{4}> -\lg 3\,$	$x>-\frac{\lg 3}{\lg \frac{5}{4}}$

Solving Inequalities

be careful with "cutting" ( $\times$ / $\div$ or $+\,$ / $-\,$ )
be VERY careful when dividing $\log\,$ (check whether it is negative)
if asked for integer, give integer
CHECK answer

n integer

If we are asked

least value of $n\,$ , we should get $n\,$ $>\,x$
largest value of $n\,$ , we should get $n\,$ $<\,x$

For example

$\,n>3.9\therefore$ least integer $n\,$ is $4\,$
$\,n>4.1\therefore$ least integer $n\,$ is $5\,$
$\,n<7.2\therefore$ largest integer $n\,$ is $7\,$
$\,n<10.9\therefore$ largest integer $n\,$ is $10\,$
$\,n<10.1\therefore$ largest integer $n\,$ is

Examples

$2^{x}>500\,$ (3 s.f.)
Analyze:
- $2^{x}>500\,$ log both sides (base ten will do)
- $\lg 2^{x}>\lg 500\,$
- $x\lg 2>\lg 500\,$ $\lg 2\,$ is positive
- $x>\frac{\lg 500}{\lg 2}\,$
- $x>8.97\mbox{(3 s.f.)}\,$

Find smallest integer $n\,$ such that $\left(0.9\right)^{n}<0.01$
Analyze :
- $\left(0.9\right)^{n}<0.01$
- $\lg\left(0.9\right)^{n}<\lg 0.01$
- $n\lg 0.9<\lg 0.01\,$ Be careful! $\lg 0.9\,$ is negative
- $n>\frac{\lg 0.01}{\lg 0.9}$
- $n>43.709\,$
- $\therefore \mbox{The smallest integer } n \mbox { is } 44$

Find largest integral value of $x\,$ such that $3\left(2^{x}\right)<5000$
Try it on your own first! Including checking the answer. The answer is $x=10\,$
- $\begin{align} 3\left(2^{x}\right)&<5000\\ 2^{x}&<\frac{5000}{3}\\ \lg 2^{x}&<\lg \left(\frac{5000}{3}\right)\\ x \lg 2&<\lg \left(\frac{5000}{3}\right)\\ x &<\frac{\lg \left(\frac{5000}{3}\right)}{\lg 2}\\ x &< 10.702 \\ \therefore& \mbox{The largest integral value of } x \mbox { is } 10 \end{align}$
  Check : $3\left(2^{10}\right)=3072, 3\left(2^{11}\right)=6144,$

Find least integral value of $x\,$ such that $6\left(\frac{2}{3}\right)^{x-1}<\frac{1}{1000}$
Analyze :
- $\begin{align} 6\left(\frac{2}{3}\right)^{x-1}&<\frac{1}{1000}\\ \left(\frac{2}{3}\right)^{x-1}&<\frac{1}{6000}\\ \lg \left(\frac{2}{3}\right)^{x-1}&<\lg \left(\frac{1}{6000}\right)\\ \left(x-1\right) \lg \left(\frac{2}{3}\right)&>\lg \left(\frac{1}{6000}\right)\\ x-1 &>\frac{\lg \left(\frac{1}{6000}\right)}{\lg \left(\frac{2}{3}\right)}\\ x-1 & > 21.456 \\ x &> 22.456 \\ \therefore& \mbox{The least integral value of } x \mbox { is } 23 \end{align}$
  Check : $6\left(\frac{2}{3}\right)^{23-1}=0.000802, 6\left(\frac{2}{3}\right)^{22-1}=0.0012$

Exercise 10

1) Find the smallest integer $n\,$ such that $3^{n}>25600\,$ $n=10\,$

- $\begin{align} 3^{n}&>25600 \\ \lg 3^{n}&>\lg 25600 \\ n \lg 3 &>\lg 25600 \\ n &>\frac{\lg 25600}{\lg 3} \\ n &>9.239\\ \therefore & \mbox{ smallest integer } n=10 \end{align}$
- Check : $3^{10}=59049, 3^9=19683\,$

2) Find the smallest integral value of $n\,$ such that $\left(0.85\right)^{n}<0.0001\,$ $n=57\,$

- $\begin{align} \left(0.85\right)^{n}&<0.0001 \\ \lg \left(0.85\right)^{n}&<\lg 0.0001 \\ n\lg 0.85&<\lg 0.0001 \\ n &>\frac{\lg 0.0001}{\lg 0.85} \\ n &>56.67\\ \therefore & \mbox{ smallest integral value of } n=57 \end{align}$
- Check : $\left(0.85\right)^{57}=0.0000948, \left(0.85\right)^{56}=0.000115\,$

3) Find the largest integral value of $n\,$ such that $7\left(4^{n}\right)<5000$ $n=4\,$

- $\begin{align} 7\left(4^{n}\right)&<5000\\ 4^{n}&<\frac{5000}{7}\\ \lg 4^{n}&<\lg \left(\frac{5000}{7}\right)\\ n\lg 4&<\lg \left(\frac{5000}{7}\right)\\ n&<\frac{\lg \left(\frac{5000}{7}\right)}{\lg 4}\\ n&<4.740\\ \therefore & \mbox{ largest integral value of } n=4 \end{align}$
- Check : $7\left(4^{4}\right)=1792, 7\left(4^{5}\right)=7168\,$

4) Find the largest integer $n\,$ such that $4\left(\frac{3}{5}\right)^{n+1}>\frac{1}{100}$ $n=10\,$

- $\begin{align} 4\left(\frac{3}{5}\right)^{n+1}&>\frac{1}{100}\\ \left(\frac{3}{5}\right)^{n+1}&>\frac{1}{400}\\ \lg \left(\frac{3}{5}\right)^{n+1}&>\lg \left(\frac{1}{400}\right)\\ \left(n+1\right)\lg \left(\frac{3}{5}\right)&>\lg \left(\frac{1}{400}\right)\\ n+1&<\frac{\lg \left(\frac{1}{400}\right)}{\lg \left(\frac{3}{5}\right)}\\ n&<10.72 \\ \therefore & \mbox{ largest integer of } n=10 \end{align}$
- Check : $4\left(\frac{3}{5}\right)^{10+1}=0.0145, 4\left(\frac{3}{5}\right)^{11+1}=0.00871\,$

Index Part2

Contents

Definition of Logarithm

Value of a & N

Special notation

Examples

Exercise 1

Laws of Logarithm

Common Mistakes

Exercise 2

Change of base

Examples

Algebra (Revision)

Exercise 3

Value of

Exercise 4

Same Base

Examples

Exercise 5

Using Substitution

Examples

Exercise 6

Using Logarithm to Solve Exponent Equations

Writing the final answer

Example

More Rearranging

Example

Identifying Type of Exponent Equations

Exercise 7

Solving Simultaneous Equations

Examples

Exercise 8

Simple Inequalities

IMPORTANT

Exercise 9

Solving Inequalities

n integer

Examples

Exercise 10

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Namespaces

Variants

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Value of $a^{\log_{a}x}$