Index Part1

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Contents

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)
    • First of all, what does \,a^{m} means?
      • Hmm... what does \,a^{3} mean? a \times a \times a
      • Thus, \,a^{m} =\underbrace{a \times a \times a \times \ldots \times a}_{m\;times}
      • And \,a^{n} =\underbrace{a \times a \times a \times \ldots \times a}_{n\;times}
    • Thus, a^{m}\times a^{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}
      • So what? Consider a^{3} \times a^{2} =\underbrace{\left(a \times a \times a\right)}_{3\;times} \times \underbrace{\left(a \times a\right)}_{2\;times}
      • The brackets here won't matter, thus =\underbrace{a\times a\times a\times a\times a}_{?\,times}
      • How many times? 3+2=5\, =\underbrace{a\times a\times a\times a\times a}_{3+2=5\,times}
      • But what does =\underbrace{a\times a\times a\times a\times a}_{5\,times} mean? =a^5\,
    • 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
    • A simpler example will be \frac{a^{5}}{a^{3}}
      • =\frac{a\times a\times a\times a\times a}{a\times a\times a}
      • Hmm... =\frac{{\color{Red}\cancel{a\times a\times a}}\times a\times a}{{\color{Red}\cancel{a\times a\times a}}} =a\times a
      • That is, \frac{a^{5}}{a^{3}}
        • =\frac{\overbrace{a\times a\times a\times a\times a}^{5\;times}}{\underbrace{a\times a\times a}_{3\;times}} =\frac{\overbrace{{\color{Red}a\times a\times a}\times \overbrace{a\times a}^{5-3=2\;times}}^{5\;times}}{\underbrace{{\color{Red}a\times a\times a}}_{3\;times}} =a\times a =a^{2}\,
    • 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^{2}\right)^{3} Equals to? a^{6}\, Urmm... we already know that. What we need to do is prove it.
    • What do we know about \left(a^{something}\right)^{something}?
      • Nothing yet. But we do know about a^{something}\,
      • In other words, we know about {\color{Red}\heartsuit}^{n}
      • Hmmm.... so how does \left(a^{2}\right)^{3} becomes {\color{Red}\heartsuit}^{n}? {\color{Red}\left(a^{2}\right)}^{3}
      • Since {\color{Red}\heartsuit}^{3}= {\color{Red}\heartsuit} \times {\color{Red}\heartsuit} \times {\color{Red}\heartsuit}
      • Thus {\color{Red}\left(a^{2}\right)}^{3}={\color{Red}\left(a^{2}\right)} \times {\color{Red}\left(a^{2}\right)} \times {\color{Red}\left(a^{2}\right)}
      • But a^{2}\,= a \times a
      • Thus,\left(a^{2}\right)^{3}
        • =a^{2}\times a^{2}\times a^{2}
        • =\left(a \times a \right) \times \left(a \times a \right) \times \left(a \times a \right) How will we know how many times? (other than counting one by one, of course
        • =\underbrace{\overbrace{\left(a \times a \right)}^{{\color{Red}2\;times}} \times \overbrace{\left(a \times a \right)}^{{\color{Red}2\;times}} \times \overbrace{\left(a \times a \right)}^{{\color{Red}2\;times}}}_{{\color{blue}3\;brackets}} =\underbrace {a \times a \times a \times a \times a \times a}_{2\times 3=6\;times} =a^{6}\,
    • \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
    • What is a^{0}\,?
      • We just know a^{m}\,, how does we make the m\, into a 0\,?
      • Hmmm... how to make 3\, into 0\,? 3-3=0\,
      • Hmmm... that means a^{3-3}\, =a^{0}\,
      • But what is a^{3-3}\,? =a^{0}\, Ermm....
      • a^{3-3}=\, \frac{a^{3}}{a^{3}} =1\,
    • 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}
    • We can then also write a more general proof if we want to

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 m\, and n\, 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}= a^{\frac{1}{2}}\, (Since \sqrt{a}=\sqrt[2]{a}\,. This should be considered a short form, that is, the 2\, should be written, but omitted just for convenience)

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

  • a^{-\frac{3}{4}} We can deal with the negative first =\frac{1}{a^{\frac{3}{4}}}
    • Then the 3\, =\frac{1}{{\color{Red}a^{3}}}, then the 4\, (in the same step) \frac{1}{{\color{Red}\sqrt[4]{{\color{Black}a^{3}}}}}
    • Or, the 4\, =\frac{1}{{\color{Red}\sqrt[4]{a}}}, then the 3\, (in the same step) \frac{1}{{\color{Red}\left(\right.}\sqrt[4]{a}{\color{Red}\left.\right)^{3}}}

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)
    • Now, we know that for multiplication, the order and position don't really matter and can be changed, so we can group all the a\,'s together and the b\,'s together
    • =ababab\,
    • =aaabbb\,
    • =a^{3}b^{3}\,
  • \left(\frac{a}{b}\right)^{n}=\frac{{a}^{n}}{{b}^{n}}
  • \left(\frac{a}{b}\right)^{3}
    • =\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}\, =5^{x}\, ? WRONG! No such thing and nothing we can do here...
  • 4\cdot 3 ^{x} =12^{x}\, ? WRONG! We would need 4^{{\color{Red}x}}\cdot 3 ^{x} to be able to do that..
  • \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 an = b

  • a^n=b, \therefore a= b^\frac{1}{n}
    • This is because we needed to change a^n\, to a^{1}\,, in other words, change n\, to 1\,
      • n \times ? =1 n \times \frac{1}{n} =1 \left(a^n\right)^\frac{1}{n}=b^\frac{1}{n}
  • a^\frac{1}{n}=b, \therefore a= 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}
  • a^{-1}=b, \therefore a= 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\,
  • \therefore x=8^{\frac{1}{3}}
    • =2\,
    • How do we know our answer is correct?
      • Firstly, what did the question ask? Value of x\, such that x^{3}=8\,
      • Does our answer satisfy the equation? 2^{3}=8\, Yes.

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

  • x^{-3}=\,8
  • x=8^{-\frac{1}{3}} =\frac{1}{2} or \frac{1}{x^{3}}=8, \therefore x^{3}=\frac{1}{8}, \therefore x=\frac{1}{2}
    • Check :\left(\frac{1}{2}\right)^{-3}=8 Any mistake will give us error on checking
  • x^{-\frac{1}{2}}=4
  • x=4^{-2}\, =\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 an

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 : First, we must learn to identify why this question is solved by this method
      • {\color{Red}4}\cdot {\color{Red}2}^{x+2}=\frac{1}{{\color{Red}8}^{x}} All base 2\,
      • {\color{Red}4\cdot 2^{x+2}}={\color{Blue}\frac{1}{8^{x}}} Two terms only
      • Thus, it can be arranged into a^{m}=a^{n}\,
    • 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
    • The most important step would be to check. The few seconds used is very much worth it.
    • 4\cdot 2^{1}=8, \frac{1}{8^{-1}}=8
  • 3\cdot 9^{x+1}-\sqrt{3^{x}}=0
    • Analyze : {\color{Red}3\cdot 9^{x+1}}-{\color{Red}\sqrt{3^{x}}}={\color{Red}0} It might look like three terms
      • but it can easily be arranged to {\color{Red}3\cdot 9^{x+1}}={\color{Red}\sqrt{3^{x}}}
    • 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 an

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 an

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 78\, is not base 3\,, and also there are more than two terms, so instead of making it {\color{Red}3}^{m}={\color{Red}3}^{n}, we try to look at {\color{Red}3}^{{\color{Red}x}+3}-{\color{Red}3^{x}}=78\,
    • 3^{x+3}-3^{x}=78\, We try to isolate the 3^{x}\,
    • 27\left(3^{x}\right)-3^{x}=78\, Now we have 27\left({\color{Red}3^{x}}\right)-{\color{Red}3^{x}}=78, it's really just 27{\color{Red}\heartsuit}-{\color{Red}\heartsuit} (You can, however also use substitution if you can't see)
    • 26\left(3^{x}\right)=78\,
    • 3^{x}=3\,
    • \therefore x=1\,
    • Remember, the most important step would be to check
      • 3^{4}-3=78\,

  • 4^{x}+7\left(2^{x}\right)-8=0
    • Analyze Both 4^{x}\, and 2^{x}\, is base 2\,, but we can't arrange to 2^{m}=2^{n}\,
    • 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
    • It is possible to do this without substitution, but for this case, it will be better if we do
    • \mbox{Let } u=2^{x}\,
    • u^{2}+7u-8=0\,
    • \left(u+8\right)\left(u-1\right)=0
    • u=-8 \mbox { or } u=1\, Remember, we need to solve for x\,, not just u\,
    • \therefore 2^{x}=-8 \mbox{ or } 2^{x}=1 Do note that 2^{x}>0\, for all values of x\,, thus we have to reject the first answer
    • \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 This is a simultaneous equation. But how do we solve this?
      • A simpler type of simultaneous equation will be something like \begin{align}& {\color{Red}a}+{\color{Blue}b}=9\\ & 2{\color{Red}a}+3{\color{Blue}b}=11\end{align}, where we can see the {\color{Red}a}'s and the {\color{Blue}b}'s
      • But now we have \begin{align}& {\color{Purple}2^{x}}+{\color{Purple}3^{y}}=11\\ & {\color{Purple}2^{x+1}}+{\color{Purple}3^{y+2}}=85\end{align}, seemingly four different things, but look carefully and we will see \begin{align}& {\color{Red}2^{x}}+{\color{Blue}3^{y}}=11\\ & {\color{Red}2^{x{\color{Black}+1}}}+{\color{Blue}3^{y{\color{Black}+2}}}=85\end{align}
    • \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}\,
    • \begin{array}{ll} a+b=11 & 2a+9b=85\frac{\qquad}{}(2) \\ 2a+2b=22\frac{\qquad}{}(1) \end{array} Any other method to solve this is fine
    • (2)-(1):\; 7b=63, \therefore b=9
    • \therefore a+9=11, \therefore a=2 As before, don't forget to solve what is asked in the question
    • \therefore 2^{x}=2, 3^{y}=9
    • \therefore x=1, y=2
    • To check, we would need to check both equations.
      • 2+3^{2}=11\,
      • 2^{2}+3^{4}=85\,

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}

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