Index Part1

From StpmWiki

Jump to: navigation, search

Notes

Even though all the laws here are the same as in SPM add maths, STPM questions will of course be more challenging. It is very important here to understand the why's and the when's of a method. Blindly memorizing, and even blindly practicing will get you nowhere.

Learning Objectives (Syllabus)

understand integral and rational exponents
understand the relationship between logarithm and exponents
carry out change of base of logarithms
use the law of exponents and laws of logarithms
use the result : for $a>b\,$ and $c>1\,$ , $c^{a}>c^{b}\,$ and $log_{c}a>log_{c}b\,$ ; for $a>b\,$ and $c<1\,$ , $c^{a}<c^{b}\,$ and $log_{c}a<log_{c}b\,$
solve equations and inequalities involving exponents and logarithms

Prior Knowledge

very basic algebraic skills, including simple simultaneous equations

Laws of Exponents

Instead of just writing the laws here, we will try to prove it. Note that the prove is not really needed in the syllabus, and the prove that we will show here isn't even really the actual mathematical prove, but it should suffice to give us an idea about the reason behind the laws. Why? As I said, we need to understand rather than blindly memorizing. In fact, trying to understand the index laws should be done during form 3 and form 4, but most teachers don't emphasize it, and thus most students can't really explain them in a convincing way.

a) $a^{m}\times a^{n}=$ $\,a^{m+n}$

Yes, simple enough, but can you explain why? (You can stop reading if you can)
- Thus $a^{m+n}\,$ $=\underbrace{\left( a \times a \times a \times \ldots \times a\right)}_{m\;times}\times \underbrace{\left(a \times a \times a \times \ldots \times a\right)}_{n\;times}$
  $=\underbrace{a \times a \times a \times \ldots \times a}_{m+n\;times}$
  $=a^{m+n}\,$

b) $\frac{a^{m}}{a^{n}}=$ $\,a^{m-n}$

Try writing out the "prove" yourself. You can always look at specific situations to give you an idea how to write a general one
- Thus $\frac{a^{m}}{a^{n}}$
  = $\frac{\overbrace{a\times a\times \ldots \times a}^{m\;times}}{\underbrace{a\times a\times \ldots \times a}_{n\;times}}$ $\frac{\overbrace{\overbrace{\cancel{a\times a\times \ldots \times a}}^{n\;times}\times \overbrace{a\times a\times \ldots \times a}^{m-n\;times} }^{m\;times}}{\underbrace{\cancel{a\times a\times \ldots \times a}}_{n\;times}}$
  $=\overbrace{a\times a\times \ldots \times a}^{m-n\;times}$ $=a^{m-n}\,$

c) $\left(a^{m}\right)^{n}=$ $\,a^{mn}$

Again, a specific example will set us on the right way
- $\left(a^{m}\right)^{n}$
  $=\underbrace{a^{m} \times a^{m} \times \ldots \times a^{m}}_{n\;times}$
  $=\underbrace{ \overbrace{\left(a \times \ldots \times a\right)}^{m\;times} \times \overbrace{\left(a \times \ldots \times a\right)}^{m\;times} \times \ldots \times \overbrace{\left(a \times \ldots \times a\right)}^{m\;times}}_{n\;brackets}$
  $=\underbrace{a \times a \times \ldots a}_{m \times n \; times}$ $=a^{mn}\,$

d) $a^{0}\,$ $=1\,$

This won't be like above, it involves a smart trick
- $a^{0}\,$ $=a^{m-m}\,$ $=\frac{a^{m}}{a^{m}}$ $=1\,$

e) $a^{-n}\,$ $=1\,$

This, again, involves another trick
- How to get negative numbers? What is $-3\,?$ $=0-3\,$
- $a^{-3}\,$ $=a^{0-3}\,$ $=\frac{a^{0}}{a^{3}}$ $=\frac{1}{a^{3}}$
- $a^{-n}\,$ $=a^{0-n}\,$ $=\frac{a^{0}}{a^{n}}$ $=\frac{1}{a^{n}}$

f) $a^{\frac{1}{n}}\,$ $=\sqrt[n]{a}$

This one require some knowledge of roots
- $\sqrt[3]{a}$ , for example, will be such that $\sqrt[3]{a} \times \sqrt[3]{a} \times \sqrt[3]{a} = a$
- Hmmmm... but $a^{?} \times a^{?} \times a^{?} = a$
- But we can see that $?+?+?=1\,$ , meaning $?=\frac{1}{3}$
- That is, $a^{\frac{1}{3}}=\sqrt[3]{a}$

g) $a^{\frac{m}{n}}$ $=\sqrt[n]{a^{m}}$ or $=\left(\sqrt[n]{a}\right)^{m}$

We need to find a way to deal with the and separately
- $\frac{m}{n}$ $=m\left(\frac{1}{n}\right)$ or $=n\left(\frac{1}{m}\right)$
- $a^{\frac{m}{n}}$ $=\left(a^{m}\right)^{\frac{1}{n}}$ or $=\left(a^{\frac{1}{n}}\right)^{m}$
  $=\sqrt[n]{a^{m}}$ or $\left(\sqrt[n]{a}\right)^{m}$

Notes

$\frac{1}{a}=$ $a^{-1}\,$ (Since $a=a^{1}\,$ )
$\sqrt{a}=$

Common Mistakes

$\frac{1}{a^{3}}=a^{-3}$ Correct
$\frac{1}{a^{3}}=a^{3}$ Wrong (Forgetting the negative)
$\sqrt[3]{a}=a^{\frac{1}{3}}$ Correct
$\sqrt[3]{a}=a^{\frac{3}{2}}$ Wrong (Misunderstanding of the $\sqrt[n]{\quad}$ ) sign
$\sqrt[3]{a^{5}}=a^{\frac{3}{5}}$ Wrong
$\sqrt[3]{a^{5}}=a^{\frac{5}{3}}$ Correct

Combination

When it involves too many things, we can always deal with one at a time

We can deal with the negative first $=\frac{1}{a^{\frac{3}{4}}}$

Exercise 1

Complete the following table

1	$\sqrt[3]{a}$	$a^{\frac{1}{3}}$	7	$\frac{1}{\left(\sqrt[5]{a}\right)^{2}}$	$a^{-\frac{2}{5}}$
2	$\sqrt{a}$	$a^{\frac{1}{2}}$	8	$\frac{1}{a}$	$a^{-1}\,$
3	$\frac{1}{a}$	$a^{-1}\,$	9	$\frac{1}{a^{5}}$	$a^{-5}\,$
4	$\frac{1}{a^{3}}$	$a^{-3}\,$	10	$\sqrt[5]{a^{3}}$	$a^{\frac{3}{5}}$
5	$\frac{1}{\sqrt[3]{a}}$	$a^{-\frac{1}{3}}$	11	$\frac{1}{\sqrt{a}}$	$a^{-\frac{1}{2}}$
6	$\sqrt[3]{a^{2}}$	$a^{\frac{2}{3}}$	12	$\frac{1}{\sqrt[3]{a^{4}}}$	$a^{-\frac{4}{3}}$

Product and Quotient

$\left(a\cdot b\right)^{n}$ $=a^{n}b^{n}\,$
$\left(\frac{a}{b}\right)^{n}$ $=\frac{{a}^{n}}{{b}^{n}}\,$

Proof

Again, the proof is not in the syllabus (in fact, most text books don't even list down the two above laws), but it is important to understand the reason behind it.

$\left(ab\right)^{n}=a^{n}b^{n}$
We shall just prove for a specific case to get an idea of the reason
- $\left(ab\right)^{3}$
- $=\left(ab\right)\times\left(ab\right)\times\left(ab\right)$
- $=ababab\,$
- $=aaabbb\,$
- $=a^{3}b^{3}\,$
$\left(\frac{a}{b}\right)^{n}=\frac{{a}^{n}}{{b}^{n}}$
- $=\left(\frac{a}{b}\right)\times\left(\frac{a}{b}\right)\times\left(\frac{a}{b}\right)$
- $=\left(\frac{aaa}{bbb}\right)$
- $=\frac{a^{3}}{b^{3}}$

Notes

It DOES NOT apply to addition/subtraction! Which of the below is correct and which is wrong?

$\left(a+b\right)^{2}=a^{2}+b^{2}$ WRONG
$\left(2+b\right)^{3}=8+b^{3}$ WRONG
$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$ WRONG
$\sqrt{9-x^{2}}=3-x$ WRONG

But why?

Look at the above prove for product and quotient
- Then if we try to do something similar for $\left(a+b\right)^{3}$
- We get $\left(a+b\right)\left(a+b\right)\left(a+b\right)$
- Which DOES NOT become $aaa+bbb\,$
- Square roots are just another type of (more complicated) $\left(\right)^{n}$ , so of course it won't work for addition/subtraction inside square root too (even though at times it is very tempting to simplify it just like that)

Combining Into a Single Exponent

Under certain condition, we would instead try to bring back the two exponents into one

$2^{x}\cdot3 ^{x}$ $=6^{x}\,$
$2^{x}+3 ^{x}\,$
$4\cdot 3 ^{x}$
$\frac{3^{x}}{2 ^{x}}$ $=\left(\frac{3}{2}\right)^{x}$

Numbers and Square roots

It is common to sometimes forget that the $a\,$ and $b\,$ can be replaced be any other unknowns, combination of unknowns, or numbers. Thus, again, we should really remember it as

$\left({\color{Red}\heartsuit}{\color{Blue}\bigcirc}\right)^{n}={\color{Red}\heartsuit}^{n}{\color{Blue}\bigcirc}^{n}$
$\left(\frac{{\color{Red}\heartsuit}}{{\color{Blue}\bigcirc}}\right)^{n}=\frac{{{\color{Red}\heartsuit}}^{n}}{{{\color{Blue}\bigcirc}}^{n}}$

Thus, if it has a mixture of unknowns and numbers, be a little more careful (or do another step) if you know that you regularly make such mistakes.

$\left(3x\right)^{2}$ $={\color{Red}3}^{2}{\color{Blue}x}^{2}= 9x^{2}\,$ and not $3x^{2}\,$ or $6x^{2}\,$
$\left(5x\right)^{-3}$ $=\frac{1}{\left(5x\right)^{3}}= \frac{1}{125x^{3}}$ and not $\frac{125}{x^{3}}\,$ or $-15x^{-2}\,$

Also, do note that this also applies for $\sqrt{\quad}$ and $\sqrt[n]{\quad}$

$\sqrt{9x^{2}}$ $=\sqrt{{\color{Red}9}}\sqrt{{\color{Blue}x^{2}}}=3x$ and not $9x\,$

Exercise 2

Complete the following table

1	$\left(2x\right)^{3}$	$8x^{3}\,$	5	$\left(\frac{5}{x}\right)^{2}$	$\frac{25}{x^{2}}$
2	$\left(3x\right)^{-4}$	$\frac{1}{81x^{4}}$	6	$\sqrt{\frac{3}{4}x^{2}}$	$\frac{\sqrt{3}}{2}x$
3	$\sqrt{4x}$	$2\sqrt{x}$	7	$\sqrt{4x}$	$2\sqrt{x}$
4	$\sqrt{3x^{2}}$	$\sqrt{3}x$	8	$\sqrt[3]{512x}$	$8\sqrt[3]{x}$

Solving $a n = b$

$b^\frac{1}{n}$
- This is because we needed to change to , in other words, change to
  - $n \times ? =1$ $n \times \frac{1}{n} =1$ $\left(a^n\right)^\frac{1}{n}=b^\frac{1}{n}$
$b^n\,$
- Similarly, we change $\frac{1}{n}\,$ to $1\,$ by $\frac{1}{n}\times n=1$
$a^{\frac{m}{n}}=b, \therefore a=$ $b^\frac{n}{m}$
$b^{-1}\,$
- This can be seen from $-1\times -1 =1$ or from $\frac{1}{a}=b$

Checking

One thing I will always repeat from now on (at least until we reach the second half of the syllabus where it becomes very difficult to verify the answers), is CHECK your answers. By checking, I don't mean turning to the back of your textbook (or clicking on the solution in our situation), but to verify that our answer does indeed satisfy the question. In a lot of cases, this simply requires a few seconds with the calculator. It is very much worth it. TRUST ME.

Examples

Find value of $x\,$

$x^{3}=8\,$
- $=2\,$

$x^{\frac{1}{3}}=8\,$
$=512\,$
- Check : $512^{\frac{1}{3}}=8$

$x^{-3}=\,8$
$=\frac{1}{2}$ or $\frac{1}{x^{3}}=8, \therefore x^{3}=\frac{1}{8}, \therefore x=\frac{1}{2}$
$x^{-\frac{1}{2}}=4$
$=\frac{1}{16}$
- Check : $\left(\frac{1}{16}\right)^{-\frac{1}{2}}=4$

Same Base

If $a^{n}=a^{m}, \left(a\neq 0,\pm1 \right)$

$\therefore n=m$

Solving Equations with Same Base

Identifying same base
- Ex: $3,9,27,\frac{1}{3}, \frac{1}{27}, \sqrt{3},\frac{1}{\sqrt[4]{81}}\rightarrow$ all base $\,3$
Arrange into $a^{m}=a^{n}, \therefore m=n$
ALWAYS check answer! By substituting back into original equation

Changing to $a n$

Example: Change the following to the form of $3^{n}\,$

$9\cdot 3^{x}$ $=3^{2}\cdot3^{x}\,$ $=3^{x+2}\,$
$3\cdot 9^{x}$ $=3\cdot 3^{2x}\,$ $=3^{2x+1}\,$
$\sqrt{3^{x}}$ $=\left(3^{x}\right)^{\frac{1}{2}}\,$ $=3^{\frac{1}{2}x}$
$9^{x+2}\,$ $=\left(3^{2}\right)^{x+2}\,$ $=3^{2\left(x+2\right)}$ $=3^{2x+4}\,$
$\frac{27^{x-1}}{9}\,$ $=\frac{3^{3x-3}}{3^{2}}$ $=3^{3x-5}\,$

Examples

Find the value of $x\,$

$4\cdot 2^{x+2}=\frac{1}{8^{x}}$
- Analyze :
- $4\cdot 2^{x+2}=\frac{1}{8^{x}}$
- $2^{2}\cdot 2^{x+2}=\frac{1}{2^{3x}}$
- $2^{x+4}=2^{-3x}\,$
- $\therefore x+4=-3x$
- $\therefore 4x=-4$
- $\therefore x=-1$
- $4\cdot 2^{1}=8, \frac{1}{8^{-1}}=8$

$3\cdot 9^{x+1}-\sqrt{3^{x}}=0$
- Analyze :
- $3\cdot 9^{x+1}-\sqrt{3^{x}}=0$
- $3\cdot 9^{x+1}=\sqrt{3^{x}}$
- $3\cdot 3^{2\left(x+1\right)}=3^{\frac{x}{2}}$
- $3^{2x+3}=3^{\frac{x}{2}}$
- $\therefore 2x+3=\frac{x}{2}$
- $\therefore \frac{3}{2}x=-3$
- $\therefore x=-2$
- Check : $3\cdot 9^{-1}-\sqrt{3^{-2}}=0$

Exercise 3

Find the value of $x\,$

1) $2^{x-3}=2 \cdot 4^{x}$ $x=-7\,$

- $\begin{align} & 2^{x-3}=2 \cdot 4^{x} \\ & 2^{x-3}=2 \cdot 2^{2x} \\ & 2^{x-3}= 2^{2x+1} \\ & \therefore x-3 = 2x+1 \\ & x=-4 \end{align}$
- Check : $2^{-7}=\frac{1}{128}, 2\cdot4^{-4}=\frac{1}{128}$

2) $\left(\frac{1}{5}\right)^{2x}=\frac{25^{x}}{5}$ $x=\frac{1}{4}\,$

- $\begin{align} & \left(\frac{1}{5}\right)^{2x}=\frac{25^{x}}{5} \\ & 5^{-2x}=\frac{5^{2x}}{5} \\ & 5^{-2x}=5^{2x-1} \\ & \therefore -2x = 2x-1 \\ & 4x=1 \\ & x=\frac{1}{4} \end{align}$
- Check : $\left(\frac{1}{5}\right)^{\frac{1}{2}}=0.44721, \frac{25^{\frac{1}{2}}}{5}=0.44721$

3) $8^{x-3}=2 \cdot 4^{x+2}$ $x=14\,$

- $\begin{align} & 8^{x-3}=2 \cdot 4^{x+2} \\ & 2^{3x-9}=2 \cdot 2^{2x+4} \\ & 2^{3x-9}=2^{2x+5} \\ & \therefore 3x-9 = 2x+5 \\ & x=14 \end{align}$
- Check : $8^{11}=8589.. , 2 \cdot 4^{16}=8589..$

4) $\left(\sqrt{8}\right)^{x}-\frac{1}{16}=0$ $x=-\frac{8}{3}$

- $\begin{align} &\left(\sqrt{8}\right)^{x}-\frac{1}{16}=0\\ & \left(\sqrt{8}\right)^{x}=\frac{1}{16}\\ &\left(\sqrt{2^{3}}\right)^{x}=\frac{1}{2^{4}}\\ & 2^{\frac{3}{2}x}=2^{-4} \\ & \therefore \frac{3}{2}x = -4 \\ & x=-\frac{8}{3} \end{align}$
- Check : $\left(\sqrt{8}\right)^{-\frac{8}{3}}-\frac{1}{16}=0$

5) $\left(\sqrt[3]{9}\right)^{x}=\frac{1}{3}$ $x=-\frac{3}{2}$

- $\begin{align} &\left(\sqrt[3]{9}\right)^{x}=\frac{1}{3}\\ & \left(3^{\frac{2}{3}}\right)^{x}=3^{-1}\\ & 3^{\frac{2}{3}x}=3^{-1} \\ & \therefore \frac{2}{3}x = -1 \\ & x=-\frac{3}{2} \end{align}$
- Check : $\left(\sqrt[3]{9}\right)^{-\frac{3}{2}}=\frac{1}{3}$

6) $2^{\left(x-1\right)^{2}}=\frac{8^{x}}{32}$ $x=2 \;\mbox{or}\; x=3$

- $\begin{align} &2^{\left(x-1\right)^{2}}=\frac{8^{x}}{32}\\ &2^{\left(x-1\right)^{2}}=\frac{2^{3x}}{2^{5}}\\ &2^{\left(x-1\right)^{2}}=2^{3x-5}\\ & \therefore \left(x-1\right)^{2} = 3x-5 \\ & x^{2}-2x+1=3x-5 \\ & x^{2}-5x+6 =0 \\ & \left(x-2\right)\left(x-3\right)=0\\ & x=2 \;\mbox{or}\; x=3 \\ \end{align}$
- Check : $2=\frac{64}{32}$ , $2^{4}=\frac{512}{32}$

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

- $\begin{align} &\left(9^{x}\right)^{x}-3\left(3\right)^{1-3x}=0\\ &\left(3^{2x}\right)^{x}=3\left(3\right)^{1-3x}\\ & 3^{2x^{2}}=3^{2-3x}\\ & \therefore 2x^{2} = 2-3x \\ & 2x^{2}+3x-2=0 \\ & \left(2x-1\right)\left(x+2\right)\\ & x=\frac{1}{2} \;\mbox{or}\; x=-2 \\ \end{align}$
- Check : $9^{\frac{1}{4}}-3\left(3\right)^{-\frac{1}{2}}=0$ , $9^{4}-3\left(3\right)^{7}=0$

8) $5^{x-3}-1=0\,$ $x=3\,$

- $\begin{align} & 5^{x-3}-1=0 \\ & 5^{x-3}=1 \\ & 5^{x-3}=5^{0}\\ & \therefore x-3 = 0 \\ & x=3 \end{align}$
- Check : $5^{3-3}=1\,$

Using Substitution

When necessary, we can use substitution $u=a^{x}\,$

Remember to solve for $x\,$ after solving $u\,$
As usual, ALWAYS check answer, by substituting $x\,$ (not $u\,$ ) into the original equation

Substituting $a n$

If $u=3^{x}\,$

$3^{x+1}\,$ $=3\left(3^{x}\right)$ $=3u\,$
$2\cdot 3^{x-2}\,$ $=2\left(\frac{3^{x}}{9}\right)$ $=\frac{2}{9}u\,$
$3^{x+2}-3^{x-1}\,$ $=9u-\frac{1}{3}u$ $=\frac{26}{3}u\,$
$3^{2x}\,$ $=\left(3^{x}\right)^{2}$ $=u^{2}\,$
$3^{-x}\,$ $=\frac{1}{3^{x}}$ $=\frac{1}{u}\,$
$9^{x}\,$ $=\left(3^{2}\right)^{x}$ $=\left(3^{x}\right)^{2}$ $=u^{2}\,$
$27^{x}\,$ $=3^{3x}\,$ $=u^{3}\,$
$\sqrt{9^{x}}$ $=\sqrt{3^{2x}}$ $=3^{x}\,$ $u\,$

Value of $a n$

If $a>0\,$ , which is usually the case here

$\therefore a^{x}\,$ $>0\,$ for all values of $x\,$
$\therefore a^{x}=0$ or negative values will have no solution
For example
- $3^{4}=81 \,$
- $3^{-2}=\frac{1}{9}$
- $3^{\frac{1}{3}}=\sqrt[3]{3}$
- So we can see that $3^{x}>0\,$ no matter what is the value of $x\,$
- Thus $3^{x}=0 \to$ no solution/impossible/rejected
- $3^{x}=-1 \to$ no solution/impossible/rejected

Examples

$3^{x+3}-3^{x}=78\,$
- Analyze
- $3^{x+3}-3^{x}=78\,$ We try to isolate the $3^{x}\,$
- $26\left(3^{x}\right)=78\,$
- $3^{x}=3\,$
- $\therefore x=1\,$

$4^{x}+7\left(2^{x}\right)-8=0$
- Analyze
- $4^{x}+7\left(2^{x}\right)-8=0$ We try to get the $2^{x}\,$
- $2^{2x}+7\left(2^{x}\right)-8=0$
- $\left(2^{x}\right)^{2}+7\left(2^{x}\right)-8=0$
- $\mbox{Let } u=2^{x}\,$
- $u^{2}+7u-8=0\,$
- $\left(u+8\right)\left(u-1\right)=0$
- $\therefore 2^{x}=-8 {\color{Red}\mbox{ (rejected) }} \mbox{ or } 2^{x}=1$
- $\therefore 2^{x}=2^{0}$
- $\therefore x = 0$
- Check! $1+7-8=0\,$

$2^{x}+3^{y}=11 \mbox{ and } 2^{x+1}+3^{y+2}=85\,$
- Analyze
- $\begin{align} 2^{x}+3^{y}=11 \qquad & 2^{x+1}+3^{y+2}=85 \\ & 2\left(2^{x}\right)+9\left(3^{y}\right)=85 \\ \end{align}$
- $\mbox{Let } a=2^{x},\;b=3^{y}\,$
- $(2)-(1):\; 7b=63, \therefore b=9$
- $\therefore 2^{x}=2, 3^{y}=9$
- $\therefore x=1, y=2$

Exercise 4

Express the following in terms of $u\,$ , where $u=2^{x}\,$ Note : Working is not needed in the cases which you can see directly

1) $2^{x+3}\,$ $=8u\,$

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

2) $2^{x-3}\,$ $=\frac{u}{8}\,$

- $\begin{align} 2^{x-3}& =\frac{2^{x}}{8}\\ & =\frac{u}{8} \end{align}$

3) $3 \cdot 2^{x-2}$ $=\frac{3}{4}u\,$

- $\begin{align} 3\cdot 2^{x-2} & = 3\cdot \frac{2^{x}}{4}\\ & = \frac{3}{4}u \end{align}$

4) $\frac{3}{5} \cdot 2^{x+2}$ $=\frac{12}{5}u\,$

- $\begin{align} \frac{3}{5} \cdot 2^{x+2} & = \frac{3}{5} \cdot 4 \left(2^{x}\right)\\ & = \frac{12}{5}u\\ \end{align}$

5) $2^{x+2}+2^{x-1}\,$ $=\frac{9}{2}u\,$

- $\begin{align} 2^{x+2}+2^{x-1}& =4\left(2^{x}\right)+\frac{2^{x}}{2}\\ & =4u+\frac{u}{2}\\ & =\frac{9}{2}u\\ \end{align}$

6) $2^{x+3}-3\cdot 2^{x+1}\,$ $=2u\,$

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

7) $2^{x+3}\cdot 2^{x-1}\,$ $=4u^{2}\,$

- $\begin{align} 2^{x+3}\cdot 2^{x-1}& =8u \cdot \frac{u}{2}\\ & = 4u^{2} \end{align}$

8) $\frac{2^{2x}}{2^{x+1}}$ $=\frac{u}{2}$

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

9) $2^{-x}\,$ $=\frac{1}{u}$

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

10) $2^{3-x}\,$ $=\frac{8}{u}$

- $\begin{align} 2^{3-x} & =\frac{8}{2^{x}}\\ & = \frac{8}{u} \end{align}$

11) $2^{2x}\,$ $=u^{2}\,$

- $\begin{align} 2^{2x} & =\left(2^{x}\right)^{2}\\ & = u^{2} \end{align}$

12) $2^{3x}\,$ $=u^{3}\,$

- $\begin{align} 2^{3x} & =\left(2^{x}\right)^{3}\\ & = u^{3} \end{align}$

13) $2^{4-2x}\,$ $=\frac{16}{u^{2}}$

- $\begin{align} 2^{4-2x} & =\frac{16}{2^{2x}}\\ & =\frac{16}{u^{2}} \end{align}$

14) $4^{x+1}\,$ $=4u^{2}\,$

- $\begin{align} 4^{x+1} & =4\left(4^{x}\right)\\ & =4\left(2^{2}\right)^{x}\\ & =4\left(2^{x}\right)^{2}\\ & = 4u^{2} \end{align}$

15) $8^{x}\,$ $=u^{3}\,$

- $\begin{align} 8^{x} & =\left(2^{3}\right)^{x}\\ & =\left(2^{x}\right)^{3}\\ & = u^{3} \end{align}$

16) $\left(\frac{1}{2}\right)^{x}\,$ $=\frac{1}{u}$

- $\begin{align} \left(\frac{1}{2}\right)^{x} & =\frac{1}{2^{x}}\\ & = \frac{1}{u} \end{align}$

17) $\sqrt{4^{x}}$ $=u\,$

- $\begin{align} \sqrt{4^{x}} & =\sqrt{2^{2x}}\\ & =2^{x}\\ & = u \end{align}$ or $\begin{align} \sqrt{4^{x}} & =\sqrt{2^{2x}}\\ & =\sqrt{u^{2}}\\ & = u \end{align}$

18) $\sqrt{4^{x+1}}$ $=2u\,$

- $\begin{align} \sqrt{4^{x+1}} & =\sqrt{4\cdot 2^{2x}}\\ & =2\cdot 2^{x}\\ & = 2u \end{align}$ or $\begin{align} \sqrt{4^{x+1}} & =\sqrt{4\cdot 2^{2x}}\\ & =\sqrt{4u^{2}}\\ & = 2u \end{align}$

Exercise 5

Solve the following equations

1) $5^{x+1}+5^{x-2}=126\,$ $x=2\;$

- $\begin{align} & 5^{x+1}+5^{x-2}=126 \\ & 5\left(5^x\right)+\frac{5^{x}}{25}=126 \\ & \frac{126}{25}\left(5^{x}\right)=126 \\ & 5^{x} = 25 \\ & 5^{x} = 5^{2} \\ & x=2 \\ \end{align}$
- Check : $5^{3}+5=126\,$

2) $\frac{4^{x-1}+3}{2-4^{x}}=-\frac{1}{2}$ $x=2\;$

- $\begin{align} & \frac{4^{x-1}+3}{2-4^{x}}=-\frac{1}{2} \\ & 2\left(4^{x-1}+3\right)=-\left(2-4^{x}\right)\\ & 2\left(\frac{4^{x}}{4}+3\right)=4^{x}-2 \\ & \frac{1}{2}\left(4^{x}+12\right)=4^{x}-2 \\ & 4^{x}+12=2\left(4^{x}\right)-4\\ & 4^{x}=16 \\ & 4^{x}=4^{2} \\ & x=2 \end{align}$
- Check : $\frac{4+3}{2-16}=\frac{7}{-14}=-\frac{1}{2}$

3) $4^{x}-5\left(2^{x-1}\right)+1=0\,$ $x=-1\;\mbox{or}\;x=1$

- $\begin{align} & 4^{x}-5\left(2^{x-1}\right)+1=0 \\ & 2^{2x}-\frac{5}{2}\left(2^{x}\right)+1=0 \\ & \left(2^{x}\right)^{2}-\frac{5}{2}\left(2^{x}\right)+1=0 \\ & \;\mbox{Let}\;u=2^{x} \\ & \therefore u^{2}-\frac{5}{2}u+1=0 \\ & 2u^{2}-5u+2=0 \\ & \left(2u-1\right)\left(u-2\right)=0\\ & u=\frac{1}{2} \;\mbox{or}\; u=2 \\ & \therefore 2^{x}=\frac{1}{2} \;\mbox{or}\; 2^{x}=2\\ & \therefore x=-1 \;\mbox{or}\; x=1\\ \end{align}$
- Check : $4-5+1=0\,$ , $\frac{1}{4}-\frac{5}{4}+1=0$

4) $2^{2x-3}+2^{x+2}-18=0\,$ $x=2\,$

- $\begin{align} & 2^{2x-3}+2^{x+2}-18=0 \\ & \frac{\left(2^{x}\right)^{2}}{8}+4\left(2^{x}\right)-18=0 \\ & \;\mbox{Let}\;u=2^{x} \\ & \therefore \frac{u^{2}}{8}+4u-18=0 \\ & u^{2}+32u-144=0 \\ & \left(u-4\right)\left(u+36\right)=0\\ & u=4 \;\mbox{or}\; u=-36 \\ & \therefore 2^{x}=4 \;\mbox{or}\; 2^{x}=-36\left(\mbox{rejected}\right) \\ & \therefore x=2\\ \end{align}$
- Check : $2+16-18=0\,$

5) $9^{x+1}+42\left(3^{x}\right)=15\,$ $x=-1\,$

- $\begin{align} & 9^{x+1}+42\left(3^{x}\right)=15 \\ & 9\left(3^{x}\right)^{2}+42\left(3^{x}\right)-15=0 \\ & \;\mbox{Let}\;u=3^{x} \\ & \therefore 9u^{2}+42u-15=0 \\ & \left(3u-1\right)\left(u+5\right)=0\\ & u=\frac{1}{3} \;\mbox{or}\; u=-5 \\ & \therefore 3^{x}=\frac{1}{3} \;\mbox{or}\; 3^{x}=-5\left(\mbox{rejected}\right) \\ & \therefore x=-1\\ \end{align}$
- Check : $1+\frac{42}{3}=15\,$

6) $4^{x-1}+2\left(2^{x-1}\right)=24\,$ $x=3\,$

- $\begin{align} & 4^{x-1}+2\left(2^{x-1}\right)=24 \\ & \frac{4^{x}}{4}+2^{x}=24 \\ & \frac{\left(2^{x}\right)^{2}}{4}+2^{x}=24 \\ & \;\mbox{Let}\;u=2^{x} \\ & \therefore \frac{u^{2}}{4}+u=24 \\ & u^{2}+4u=96 \\ & u^{2}+4u-96=0 \\ & \left(u-8\right)\left(u+12\right)=0\\ & u=8 \;\mbox{or}\; u=-12 \\ & \therefore 2^{x}=8 \;\mbox{or}\; 2^{x}=-12 \left(\mbox{rejected}\right)\\ & \therefore 2^{x}=2^{3}\\ & \therefore x=3\\ \end{align}$
- Check : $16+2\left(4\right)=124\,$

7) $2^{x}+2^{4-x}=10\,$ $x=1 \;\mbox{or}\; x=3$

- $\begin{align} & 2^{x}+2^{4-x}=10 \\ & 2^{x}+\frac{16}{2^{x}}=10 \\ & \;\mbox{Let}\;u=2^{x} \\ & u+\frac{16}{u}=10 \\ & u^{2}+16=10u \\ & u^{2}-10u+16=0 \\ & \left(u-2\right)\left(u-8\right)=0\\ & u=2 \;\mbox{or}\; u=8 \\ & \therefore 2^{x}=2 \;\mbox{or}\; 2^{x}=8\\ & \therefore x=1\;\mbox{or}\; x=3\\ \end{align}$
- Check : $2+8=10\,$ , $8+2=10\,$

8) $\frac{3^{x}+3^{-x}}{3^{x}+11\left(3^{-x}\right)}=\frac{1}{2}$ $x=1 \;\mbox{or}\; x=3$

- $\begin{align} & \frac{3^{x}+3^{-x}}{3^{x}+11\left(3^{-x}\right)}=\frac{1}{2}\\ & 2\left(3^{x}+3^{-x}\right)=3^{x}+11\left(3^{-x}\right)\\ & 2\left(3^{x}+\frac{1}{3^{x}}\right)=3^{x}+\frac{11}{3^{x}}\\ & \;\mbox{Let}\;u=3^{x} \\ & 2\left(u+\frac{1}{u}\right)=u+\frac{11}{u}\\ & 2u+\frac{2}{u}=u+\frac{11}{u}\\ & u=\frac{9}{u}\\ & u^{2}=9 \\ & u = \pm 3 \\ & u=3 \;\mbox{or}\; u=-3 \\ & \therefore 3^{x}=3 \;\mbox{or}\; 3^{x}=-3 \;\left(\mbox{rejected}\right)\\ & \therefore x=1 \end{align}$
- Check : $\frac{3+\frac{1}{3}}{3+\frac{11}{3}}=\frac{1}{2}$

9) $2^{x+1}=11-3^{y-1}\;\mbox{and}\;2^{x-2}+3^{y+2}=82$ $x=2 \;\mbox{and}\; y=2$

- $\begin{array}{ll} 2^{x+1}=11-3^{y-1} & 2^{x-2}+3^{y+2}=82\\ 2\left(2^{x}\right)=11-\frac{3^{y}}{3} & \frac{2^{x}}{4}+9\left(3^{y}\right)=82\\ \end{array}$
- $\mbox{Let}\;a=2^{x},b=3^{y}$
- $\begin{array}{ll} 2a=11-\frac{b}{3} & \frac{a}{4}+9b=82\\ 6a=33-b & a+36b=328 \frac{\quad}{}(2)\\ b=33-6a\frac{\quad}{}(1) \end{array}$
- $\begin{align} & (1) \to (2): \\ & a+36\left(33-6a\right)=328 \\ & a + 1188 -216a=328 \\ & 860 = 215a \\ & \therefore a = 4, \;\therefore b=33-6\left(4\right) =9 \\ & \therefore 2^{x}=4 \;\mbox{and}\;3^{y}=9 \\ & 2^{x}=2^{2} \;\mbox{and}\;3^{y}=3^{2} \\ & \therefore x=2 \;\mbox{and}\;y=2 \\ \end{align}$
- Check : $8=11-3\,$ , $1+81=82\,$

10) $\frac{5^{y+1}}{4^{x-1}}=1\;\mbox{and}\;5^{y}+2^{2x}=\frac{21}{5}$ $x=1 \;\mbox{and}\; y=-1$

- $\begin{array}{ll} \frac{5^{y+1}}{4^{x-1}}=1 & 5^{y}+2^{2x}=\frac{21}{5}\\ 5^{y+1}=4^{x-1} & 5^{y}+4^{x}=\frac{21}{5}\\ 5\left(5^{y}\right)=\frac{4^{x}}{4}\\ \end{array}$
- $\mbox{Let}\;a=5^{y},b=4^{x}$
- $\begin{array}{ll} 5a=\frac{b}{4} & a+b=\frac{21}{5}\frac{\quad}{}(2)\\ b=20a\frac{\quad}{}(1)\\ \end{array}$
- $\begin{align} & (1) \to (2): \\ & a+20a=\frac{21}{5} \\ & 21a =\frac{21}{5} \\ & a = \frac{1}{5} \\ & \therefore b = 20\left(\frac{1}{5}\right)=4 \\ & \therefore 5^{y}=\frac{1}{5} \;\mbox{and}\;4^{x}=4 \\ & \therefore y=-1 \;\mbox{and}\;x=1 \\ \end{align}$
- Check : $\frac{1}{1}=1\,$ , $\frac{1}{5}+4=\frac{21}{5}$

Index Part1

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Laws of Exponents

Notes

Common Mistakes

Combination

Exercise 1

Product and Quotient

Proof

Notes

Combining Into a Single Exponent

Numbers and Square roots

Exercise 2

Solving $a n = b$

Checking

Examples

Same Base

Solving Equations with Same Base

Changing to $a n$

Examples

Exercise 3

Using Substitution

Substituting $a n$

Value of $a n$

Examples

Exercise 4

Exercise 5

Personal tools

Namespaces

Variants

Views

Actions

Search

Navigation

Toolbox

Index Part1

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Laws of Exponents

Notes

Common Mistakes

Combination

Exercise 1

Product and Quotient

Proof

Notes

Combining Into a Single Exponent

Numbers and Square roots

Exercise 2

Solving an = b

Checking

Examples

Same Base

Solving Equations with Same Base

Changing to an

Examples

Exercise 3

Using Substitution

Substituting an

Value of an

Examples

Exercise 4

Exercise 5

Personal tools

Namespaces

Variants

Views

Actions

Search

Navigation

Toolbox

Solving $a n = b$

Changing to $a n$

Substituting $a n$

Value of $a n$