Solved Examples on Exponents and Logarithms

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve each question.
Show all work.
Use at least two methods as necessary.
Check your solutions as applicable.

Depending on time, it is highly recommended you check your work (equations) even if the question did not require it.
By checking your work, you can tell whether your answer is correct or incorrect [if the Left Hand Side (LHS) is equal or not equal to the Right Hand Side (RHS)].
Please make sure you check with the original equation rather than the modified equation. This is because the modified equation could be wrong.

(1.) NSC Simplify the following:

(1.1) $3(2x)^0$

(1.2) $\log_{(-10)}$

(1.3) $\dfrac{5^{2n + 1} + 4 * 5^{2n}}{25^n}$

$ (1.1) \\[3ex] 3(2x)^0 \\[3ex] = 3 * 1 \\[3ex] = 3 \\[5ex] (1.2) \\[3ex] \log_{(-10)} \;\;DNE \;(does\;\;not\;\;exist) \\[3ex] ...logarithm\;\;of\;\;a\;\;non-positive\;\;number\;\;DNE \\[5ex] (1.3) \\[3ex] \dfrac{5^{2n + 1} + 4 * 5^{2n}}{25^n} \\[5ex] \dfrac{5^{2n} * 5^1 + 5^{2n} * 4}{5^{2(n)}} \\[5ex] \dfrac{5^{2n} * 5 + 5^{2n} * 4}{5^{2n}} \\[5ex] \dfrac{5^{2n}(5 + 4)}{5^{2n}} \\[5ex] 5 + 4 \\[3ex] 9 $

(2.) Edexcel GCE Solve

(a) $5^x = 8$, giving your answer to 3 significant figures,

(b) $\log_2(x + 1) - \log_2{x} = \log_2{7}$

$ (a) \\[3ex] 5^x = 8 \\[3ex] \ln 5^x = \ln 8 \\[3ex] x \ln 5 = \ln 8 \\[3ex] x = \dfrac{\ln 8}{\ln 5} \\[5ex] x = \dfrac{2.079441542}{1.609437912} \\[5ex] x = 1.292029674 \\[3ex] x \approx 1.29 ...to\;\;3\;\;s.f \\[3ex] $ Check

$x = 1.292029674$
LHS	RHS
$ 5^x \\[3ex] 5^{1.292029674} \\[3ex] 8.000000003 $	$8$

$ (b) \\[3ex] \log_2(x + 1) - \log_2{x} = \log_2{7} \\[3ex] \log_2\left(\dfrac{x + 1}{x}\right) = \log_2{7} \\[5ex] same\;\;log\;\;base;\;\;equate\;\;terms \\[3ex] \dfrac{x + 1}{x} = 7 \\[5ex] x + 1 = 7x \\[3ex] 7x = x + 1 \\[3ex] 7x - x = 1 \\[3ex] 6x = 1 \\[3ex] x = \dfrac{1}{6} \\[5ex] x = 0.1666666667 \\[3ex] x \approx 0.167 \\[3ex] $ Check

$x = \dfrac{1}{6}$
LHS	RHS
$ \log_2(x + 1) - \log_2{x} \\[3ex] \log_2\left(\dfrac{1}{6} + 1\right) - \log_2\left(\dfrac{1}{6}\right) \\[5ex] \log_2\left(\dfrac{1}{6} + \dfrac{6}{6}\right) - \log_2\left(\dfrac{1}{6}\right) \\[5ex] \log_2\left(\dfrac{7}{6}\right) - \log_2\left(\dfrac{1}{6}\right) \\[5ex] \log_2\left(\dfrac{7}{6} \div \dfrac{1}{6}\right) \\[5ex] \log_2\left(\dfrac{7}{6} * \dfrac{6}{1}\right) \\[5ex] \log_2{7} $	$\log_2{7}$

(3.) ACT If $\log_3{2} = p$ and $\log_3{5} = q$, which of the following expressions is equal to $10$?

$ F.\:\: 3^{p + q} \\[3ex] G.\:\: 3^p + 3^q \\[3ex] H.\:\: 9^{p + q} \\[3ex] J.\:\: pq \\[3ex] K.\:\: p + q \\[3ex] $

$ \log_3{2} = p \\[3ex] 2 = 3^p ...Relationship \\[3ex] \log_3{5} = q \\[3ex] 5 = 3^q ...Relationship \\[3ex] 2 * 5 = 10 \\[3ex] \implies 10 = 3^p * 3^q \\[3ex] 3^p * 3^q = 3^{p + q} ...Law\:\: 1...Exp $

(4.) Edexcel GCE Solve

(a) $3^x = 5$, giving your answer to 3 significant figures,

(b) $\log_2(2x + 1) - \log_2{x} = 2$

$ (a) \\[3ex] 3^x = 5 \\[3ex] \ln 3^x = \ln 5 \\[3ex] x \ln 3 = \ln 5 \\[3ex] x = \dfrac{\ln 5}{\ln 3} \\[5ex] x = \dfrac{1.609437912}{1.098612289} \\[5ex] x = 1.46497352 \\[3ex] x \approx 1.46 ...to\;\;3\;\;s.f \\[3ex] $ Check

$x = 1.46497352$
LHS	RHS
$ 3^x \\[3ex] 3^{1.46497352} \\[3ex] 4.999999998 $	$5$

$ (b) \\[3ex] \log_2(2x + 1) - \log_2{x} = 2 \\[3ex] \log_2(2x + 1) - \log_2{x} = \log_2{2^2} \\[3ex] \log_2(2x + 1) - \log_2{x} = \log_2{4} \\[3ex] \log_2\left(\dfrac{2x + 1}{x}\right) = \log_2{4} \\[5ex] same\;\;log\;\;base;\;\;equate\;\;terms \\[3ex] \dfrac{2x + 1}{x} = 4 \\[5ex] 2x + 1 = 4x \\[3ex] 4x = 2x + 1 \\[3ex] 4x - 2x = 1 \\[3ex] 2x = 1 \\[3ex] x = \dfrac{1}{2} \\[5ex] x = 0.5 \\[3ex] $ Check

$x = \dfrac{1}{2}$
LHS	RHS
$ \log_2(2x + 1) - \log_2{x} \\[3ex] \log_2\left(2\dfrac{1}{2} + 1\right) - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2{1 + 1} - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2{2} - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2\left(2 \div \dfrac{1}{2}\right) \\[5ex] \log_2\left(2 * \dfrac{2}{1}\right) \\[5ex] \log_2{4} \\[3ex] \log_2{2^2} \\[3ex] 2\log_2{2} \\[3ex] 2(1) \\[3ex] 2 $	$2$

(6.) Edexcel GCE (a) Find, to 3 significant figures, the value of x for which $8^x = 0.8$

(b) Solve the equation
$2\log_3{x} - \log_3{7x} = 1$

$ (a) \\[3ex] 8^x = 0.8 \\[3ex] \ln 8^x = \ln 0.8 \\[3ex] x \ln 8 = \ln 0.8 \\[3ex] x = \dfrac{\ln 0.8}{\ln 8} \\[5ex] x = \dfrac{1.609437912}{1.098612289} \\[5ex] x = 1.46497352 \\[3ex] x \approx 1.46 ...to\;\;3\;\;s.f \\[3ex] $ Check

$x = 1.46497352$
LHS	RHS
$ 3^x \\[3ex] 3^{1.46497352} \\[3ex] 4.999999998 $	$5$

$x = \dfrac{1}{2}$
LHS	RHS
$ \log_2(2x + 1) - \log_2{x} \\[3ex] \log_2\left(2\dfrac{1}{2} + 1\right) - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2{1 + 1} - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2{2} - \log_2\left(\dfrac{1}{2}\right) \\[5ex] \log_2\left(2 \div \dfrac{1}{2}\right) \\[5ex] \log_2\left(2 * \dfrac{2}{1}\right) \\[5ex] \log_2{4} \\[3ex] \log_2{2^2} \\[3ex] 2\log_2{2} \\[3ex] 2(1) \\[3ex] 2 $	$2$

(15.) Use the one-to-one property of logarithms to find an exact solution for $\ln(3) + \ln(3x^2 - 5) = \ln(154)$

$ \ln(3) + \ln(3x^2 - 5) = \ln(154) \\[3ex] \ln[3(3x^2 - 5)] = \ln 154...Law\;1...Log \\[3ex] Same\;\;base \\[3ex] Equate\;\;the\;\;terms \\[3ex] 3(3x^2 - 5) = 154 \\[3ex] 9x^2 - 15 = 154 \\[3ex] 9x^2 = 154 + 15 \\[3ex] 9x^2 = 169 \\[3ex] x^2 = \dfrac{169}{9} \\[5ex] x = \pm \sqrt{\dfrac{169}{9}} \\[5ex] x = \pm \dfrac{13}{3} \\[5ex] $ Check

$x = \pm \dfrac{13}{3}$
LHS	RHS
$ \ln(3) + \ln(3x^2 - 5) \\[3ex] When\;\;x = -\dfrac{13}{3} \\[5ex] \ln(3) + \ln\left[3\left(-\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[3\left(\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[3 * \dfrac{13}{3} * \dfrac{13}{3} - 5\right] \\[5ex] \ln(3) + \ln\left[\dfrac{169}{3} - 5\right] \\[5ex] \ln\left[3\left(\dfrac{169}{3} - 5\right)\right]...Law\;1...Log \\[5ex] \ln(169 - 15) \\[3ex] \ln(154) $ $ When\;\;x = \dfrac{13}{3} \\[5ex] \ln(3) + \ln\left[3\left(\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[3 * \dfrac{13}{3} * \dfrac{13}{3} - 5\right] \\[5ex] \ln(3) + \ln\left[\dfrac{169}{3} - 5\right] \\[5ex] \ln\left[3\left(\dfrac{169}{3} - 5\right)\right]...Law\;1...Log \\[5ex] \ln(169 - 15) \\[3ex] \ln(154) $	$\ln 154$

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