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# Solved Examples on Logarithmic Expressions

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 any wrong answer.

Solve each logarithmic expression.
State the reasons(laws) for each step as applicable.
Show all work. Unless asked or implied, determine the exact values.
Unless asked or implies, do not use a calculator to solve.
Use several methods as necessary.
Check your solutions with a calculator as applicable.

For Questions $1$ through $6$
Calculate the exact values of the logarithms using only what you were given.
Show all work.
When applicable, round all answers to three decimal places.
Give reasons for each step.
Given that:

$\log_c{2} = 0.693 \\[3ex] \log_c{3} = 0.477 \\[3ex] \log_c{5} = 0.699 \\[3ex] \log_c{6} = 1.792 \\[3ex] \log_c{7} = 1.946 \\[3ex] \log_c{8} = 2.079 \\[3ex] \log_c{10} = 2.303 \\[3ex]$

(1.) $\log_c{\dfrac{6}{7}}$

$\log_c{\dfrac{6}{7}} \\[3ex] = \log_c{6} - \log_c{7} ...Law\: 2 ...Log \\[3ex] = 1.792 - 1.946 \\[3ex] = -0.154$
(2.) $\log_c{125}$

$\log_c{125} \\[3ex] = \log_c{5^3} \\[3ex] = 3\log_c{5} ...Law\: 5 ...Log \\[3ex] = 3 * 0.699 \\[3ex] = 2.097$
(3.) $\dfrac{\log_c{6}}{\log_c{7}}$

$\dfrac{\log_c{6}}{\log_c{7}} \\[3ex] = \log_c{6} \div \log_c{7} ...Law\: NA \\[3ex] = \dfrac{1.792}{1.946} \\[3ex] = 0.921$
(4.) $\log_c{75}$

$\log_c{75} \\[3ex] = \log_c{(25 * 3)} \\[3ex] = \log_c{25} + \log_c{3} ...Law\: 1 ...Log \\[3ex] \log_c{25} = \log_c{5^2} = 2\log_c{5} ...Law\: 5 ...Log \\[3ex] = 2 * 0.699 + 0.477 \\[3ex] = 1.398 + 0.477 \\[3ex] = 1.875$
(5.) $\log_c{\dfrac{1}{16}}$

$\log_c{\dfrac{1}{16}} \\[3ex] = \log_c{1} - \log_c{16} ... Law\: 2 ...Log \\[3ex] \log_c{1} = 0 ... Law\: 3 ...Log \\[3ex] \log_c{16} = \log_c{2^4} = 4 * \log_c{2} ... Law\: 5 ...Log \\[3ex] \log_c{16} = 4 * 0.693 = 2.772 \\[3ex] OR \\[3ex] \log_c{16} = \log_c{(2 * 8)} = \log_c{2} + \log_c{8} ... Law\: 1 ...Log \\[3ex] \log_c{16} = 0.693 + 2.079 = 2.772 \\[3ex] = 0 - 2.772 \\[3ex] = -2.772$
(6.) $\log_c{\dfrac{21}{c}}$

$\log_c{\dfrac{21}{c}} \\[3ex] = \log_c{21} - \log_c{c} ... Law\: 2 ...Log \\[3ex] \log_c{21} = \log_c{(3 * 7)} = \log_c{3} + \log_c{7} ... Law\: 1 ...Log \\[3ex] \log_c{21} = 0.477 + 1.946 = 2.423 \\[3ex] \log_c{c} = 1 ... Law\: 4 ...Log \\[3ex] = 2.423 - 1 \\[3ex] = 1.423$
(7.) $\log_3{3^{40}}$

$\log_3{3^{40}} \\[3ex] = 40 * \log_3{3} ...Law\: 5 ...Log \\[3ex] \log_3{3} = 1 ...Law\: 4 ...Log \\[3ex] = 40 * 1 \\[3ex] = 40$
(8.) $\log{100} - \log{10}$

We can solve this in two ways.
First Method $\log{100} - \log{10} \\[3ex] = \log{\dfrac{100}{10}} \\[3ex] = \log{10} \\[3ex] = \log_{10}{10} \\[3ex] = 1 ...Law\: 4 ...Log$

Second Method $\log{100} - \log{10} \\[3ex] \log{10} = 1 ...Law\: 4 ...Log \\[3ex] \log{100} = \log{10^2} = 2 * \log{10} ...Law\: 5 ...Log \\[3ex] \log{100} = 2 * 1 = 2\\[3ex] = 2 - 1 \\[3ex] = 1$
(9.) Simplify

$\log_{11}{12} * \log_{12}{13} * \log_{13}{14} * ... * \log_{1329}{1330} * \log_{1330}{1331}$

Let us write the part before the ellipsis (...)
$\log_{11}{12} * \log_{12}{13} * \log_{13}{14}$

Assume the trend continues, we will have...
$\log_{14}{15} * \log_{15}{16} * \log_{16}{17} ...$

We have to stop :-)
Let us handle this, two terms at a time

$\log_{11}{12} * \log_{12}{13} = \log_{11}{13} ... Law\: 6 ... Log \\[3ex] \log_{11}{13} * \log_{13}{14} = \log_{11}{14} ... Law\: 6 ... Log \\[3ex] \log_{11}{14} * \log_{14}{15} = \log_{11}{15} ... Law\: 6 ... Log \\[3ex] \log_{11}{15} * \log_{15}{16} = \log_{11}{16} ... Law\: 6 ... Log \\[3ex] \log_{11}{16} * \log_{16}{17} = \log_{11}{17} ... Law\: 6 ... Log \\[3ex]$

We have to stop again :-)
We know that $11$ will be the base.
And the number will be the number preceeding the last term
Do you understand what I mean
Let us write the last three terms so it makes more sense to you

$\log_{11}{1329} * \log_{1329}{1330} * \log_{1330}{1331} \\[3ex] \log_{11}{1329} * \log_{1329}{1330} = \log_{11}{1330} ... Law\: 6 ... Log \\[3ex] \log_{11}{1330} * \log_{1330}{1331} = \log_{11}{1331} ... Law\: 6 ... Log \\[3ex] \log_{11}{1331} = \log_{11}{11^3} \\[3ex] = 3 * \log_{11}{11} ... Law\: 5 ... Log \\[3ex] \log_{11}{11} = 1 ... Law\: 4 ... Log \\[3ex] = 3 * 1 \\[3ex] = 3$
(10.) $If \log_4{M} = \dfrac{\log_6{9}}{\log_6{4}}$

Determine the value of M

$\log_4{M} = \dfrac{\log_6{9}}{\log_6{4}}$

Multiply both sides by $\log_6{4}$
$\log_4{M} * \log_4{6} = \log_6{9}$

$\log_4{M} * \log_4{6} = \log_6{9} \\[3ex] \log_4{M} * \log_4{6} = \log_6{M} ...Law\: 6 ...Log \\[3ex] \log_6{M} = \log_6{9} \\[3ex] M = 9$
(11.) $\log_p{p^{7}}$

$\log_p{p^{7}} \\[3ex] = 7 * \log_p{p} ...Law\: 5 ...Log \\[3ex] \log_p{p} = 1 ...Law\: 4 ...Log \\[3ex] = 7 * 1 \\[3ex] = 7$
(12.) $7^{\log_{7}{(3x + 10)}}$

$7^{\log_{7}{(3x + 10)}} \\[3ex] = 3x + 10 ...Law\: 7 ...Log$
(13.) $\ln e^{12}$

$\ln e^{12} \\[3ex] = \log_e{e^{12}} \\[3ex] = 12 * \log_e{e} ...Law\: 5 ...Log \\[3ex] \log_e{e} = 1 ...Law\: 4 ...Log \\[3ex] = 12 * 1 \\[3ex] = 12$
(14.) $e^{\ln{c}}$

$e^{\ln{c}} \\[3ex] = e^{\log_e{c}} \\[3ex] = c ...Law\: 7 ...Log$
(15.) Given: $\log_c{e}= 3; \log_c{f} = 4; \log_c{g} = 5$

Calculate: $\log_c{\dfrac{\sqrt[4]{f^3g^4}}{\sqrt[4]{e^4g^{-4}}}}$

$\log_c{\dfrac{\sqrt[4]{f^3g^4}}{\sqrt[4]{e^3g^{-4}}}} \\[5ex] = \log_c{\sqrt[4]{f^3g^4}} - \log_c{\sqrt[4]{e^3g^{-4}}} ...Law\: 2 ...Log$
$\log_c{\sqrt[4]{f^3g^4}} = (f^3g^4)$${\dfrac{1}{4}} ... Law 7 ... Exp (f^3g^4)$${\dfrac{1}{4}}$ = $f$$3 * {\dfrac{1}{4}} * g$$4 * {\dfrac{1}{4}}$ ... Law 5 ... Exp

$(f^3g^4)$${\dfrac{1}{4}} = f$${\dfrac{3}{4}}$ * $g^1$

$\log_c{\sqrt[4]{e^3g^{-4}}} = (e^3g^{-4})$${\dfrac{1}{4}} ... Law 7 ... Exp (e^4g^{-4})$${\dfrac{1}{4}}$ = $e$$4 * {\dfrac{1}{4}} * g$$-4 * {\dfrac{1}{4}}$ ... Law 5 ... Exp

$(e^4g^{-4})$${\dfrac{1}{4}} = e^1 * g^{-1} = \log_c{(f^{\dfrac{3}{4}} * g^1)} - \log_c{(e^{\dfrac{3}{4}} * g^{-1})} \\[5ex] \log_c{(f^{\dfrac{3}{4}} * g^1)} = \log_c{f^{\dfrac{3}{4}}} + \log_c{g^1} ... Law\: 1 ... Log \\[5ex] \log_c{(f^{\dfrac{3}{4}} * g^1)} = \dfrac{3}{4} * \log_c{f} + \log_c{g} ... Law\: 5 ... Log \\[5ex] \log_c{(f^{\dfrac{3}{4}} * g^1)} = \dfrac{3}{4} * 4 + 5 \\[3ex] \log_c{(f^{\dfrac{3}{4}} * g^1)} = 3 + 5 = 8 \\[5ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = \log_c{e^{\dfrac{3}{4}}} + \log_c{g^{-1}} ... Law\: 1 ... Log \\[3ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = \log_c{e} + -1 * \log_c{g} ... Law\: 5 ... Log \\[3ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = 3 + -1 * 5 \\[3ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = 3 + -5 \\[3ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = 3 - 5 \\[3ex] \log_c{(e^{\dfrac{3}{4}} * g^{-1})} = -2 \\[5ex] = 8 - (-2) \\[3ex] = 8 + 2 \\[3ex] = 10 (16.) \log_e{(e^{|10x - 3|})} \log_e{(e^{|10x - 3|})} \\[3ex] = |10x - 3| * \log_e{e} ...Law\: 5 ...Log \\[3ex] \log_e{e} = 1 ...Law\: 4 ...Log \\[3ex] = |10x - 3| * 1 \\[3ex] = |10x - 3| (17.) Simplify \log_8{\sqrt[3]{64}} We shall solve this question in two ways. First Method \log_8{\sqrt[3]{64}} \\[3ex] = \log_8{64^{\dfrac{1}{3}}} ... Law\: 7 ... Exp \\[5ex] = \log_8{8^{2 \left(\dfrac{1}{3}\right)}} \\[5ex] = \log_8{8^{\dfrac{2}{3}}} ... Law\: 5 ... Exp \\[5ex] = \dfrac{2}{3} * \log_8{8} \\[5ex] \log_8{8} = 1 ... Law\: 4 ... Log \\[3ex] = \dfrac{2}{3} * 1 \\[5ex] = \dfrac{2}{3} Second Method Change of Base of Logarithm \sqrt[3]{64} = 4 It is better to change it to base 2. This is because 2^2 = 4 and 2^3 = 8 You should not use base 4 because 4^2 \ne 8 \log_8{\sqrt[3]{64}} \\[3ex] = \log_8{4} \\[3ex] = \dfrac{\log_2{4}}{\log_2{8}} ...Law\: 6 ... Log \\[5ex] \log_2{4} = \log_2{2^2} = 2 * \log_2{2} ... Law\: 5 ... Log \\[3ex] \log_2{2} = 1 ... Law\: 4 ... Log \\[3ex] \log_2{4} = 2 * 1 = 2 \\[3ex] \log_2{8} = \log_2{2^3} = 3 * \log_2{2} ... Law\: 5 ... Log \\[3ex] \log_2{8} = 3 * 1 = 3 \\[3ex] = \dfrac{2}{3} (18.) \ln \sqrt[7]{e} \ln \sqrt[7]{e} \\[3ex] = \log_e \sqrt[7]{e} \\[3ex] \sqrt[7]{e} = e^{\dfrac{1}{7}} ... Law\: 7 ...Exp \\[5ex] = \log_e{e^{\dfrac{1}{7}}} \\[5ex] = \dfrac{1}{7} * log_e{e} ... Law\: 5 ... Log \\[5ex] \log_e{e} = 1 ... Law\: 4 ... Log \\[3ex] = \dfrac{1}{7} * 1 \\[5ex] = \dfrac{1}{7} (19.) Simplify 7$$\log_7{4} - \log_7{12}$

$7$$\log_7{4} - \log_7{12} \log_7{4} - \log_7{12} = \log_7{\dfrac{4}{12}} ... Law\: 2 ... Log \\[5ex] = 7$$\log_7{\dfrac{4}{12}}$ $= \dfrac{4}{12} ... Law\: 7 ... Log \\[5ex] = \dfrac{1}{3}$
(20.) $e^{\log_{e^2}{144}}$

$e^{\log_{e^2}{144}}$

$\log_{e^2}{144} \\[2ex] = \dfrac{\log_e{144}}{\log_e{e^2}} ... Law\: 6 ... Log$

$\log_e{144} \\[2ex] = \log_e{12^2} \\[2ex] = 2 * \log_e{12} ... Law\: 5 ... Log$

$\log_e{e^2} \\[2ex] = 2 * \log_e{e} ... Law\: 5 ... Log \\[2ex] = 2$

$\log_{e^2}{144} \\[2ex] = \dfrac{\log_e{144}}{\log_e{e^2}} ... Law\: 6 ... Log \\[5ex] = \dfrac{2 * \log_e{12}}{2} \\[5ex] = \log_e{12}$

$e^{\log_{e^2}{144}} \\[3ex] = e^{\log_e{12}} = 12 ... Law\: 7 ... Log$

(21.) Given that $c = \log_8{81}$ and $d = \log_2{9}$
(a.) Express $c$ in terms of $d$
(b.) Express $d$ in terms of $c$

We have two bases - base $8$ for $c$ and base $2$ for $d$
Let us express the larger base in terms of the smaller base
This means that we have to convert base $8$ to base $2$

$c = \log_8{81} \\[3ex] \log_8{81} = \dfrac{\log_2{81}}{\log_2{8}} ...Law\: 6...Log \\[5ex] \log_2{8} = \log_2{2^3} = 3\log_2{2} ...Law\: 5...Log \\[3ex] \log_2{2} = 1 ...Law\: 4...Log \\[3ex] 3\log_2{2} = 3 * 1 = 3 \\[3ex] \log_2{81} = \log_2{9 * 9} \\[3ex] \log_2{9 * 9} = \log_2{9} + \log_2{9} ...Law\: 1...Log \\[3ex] \log_2{9} + \log_2{9} = d + d = 2d ...Substitution \\[3ex] \rightarrow \log_8{81} = \dfrac{2d}{3} \\[5ex] c = \dfrac{2d}{3} \\[3ex] 3c = 2d \\[3ex] 2d = 3c \\[3ex] d = \dfrac{3c}{2}$
(22.) ACT What integer does $3(\log_2{16})$ equal?

We can do this in two ways
I think it is faster you choose the first method
However, you can solve it using the second method.
Because the ACT is a timed test, you do not need to show all work like I did.
I had to show my work to make sure "all" my students understand

$\underline{First\:\: Method}: Use\:\: TI-84/84\: Plus\:\: Calculator \\[3ex] See\:\: link\:\: on\:\: main\:\: page \\[3ex] \underline{Second\:\: Method}: Law\:\: of\:\: Logarithms \\[3ex] 3(\log_2{16}) \\[3ex] \log_2{16} = \log_2{2^4} = 4\log_2{2} ...Law\: 5...Log \\[3ex] \log_2{2} = 1 ...Law\: 4...Log \\[3ex] \implies \log_2{16} = 4 * 1 = 4 \\[3ex] = 3 * 4 \\[3ex] = 12$
(23.) JAMB Given that $\log 2 = 0.3010$, $\log 7 = 0.8451$, evaluate $\log 112$.

Because JAMB is a timed exam and we cannot use a calculator, it is better to just divide $112$ by $7$

$\log 112 = \log (16 * 7) \\[3ex] = \log(2^4 * 7) \\[3ex] = \log 2^4 + \log 7 ...Law\:\: 1...Log \\[3ex] \log 2^4 = 4 * \log 2 ...Law\:\: 5...Log \\[3ex] = 4 * \log 2 + \log 7 \\[3ex] = 4(0.3010) + 0.8451 \\[3ex] = 1.204 + 0.8451 \\[3ex] = 2.0491$
(24.) WASSCE Using $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$, evaluate $\log_{10} 0.24$

We want to express $0.24$ using only what we are given and probably what we should know.
We notice that the logarithm is in base $10$
Let us see if we can get rid of the decimal first
$0.24 = \dfrac{24}{100}$

Then, we can express $24$ in terms of $2's$ and $3's$
$24 = 8 * 3 = 2 * 2 * 2 * 3 = 2^3 * 3$

$\log_{10}{0.24} = \log_{10} \left(\dfrac{24}{100} \right) \\[5ex] \log_{10} \left(\dfrac{24}{100} \right) = \log_{10}{24} - \log_{10}{100} ...Law\:\: 2...Log \\[3ex] \log_{10}{100} = \log_{10}{10^2} = 2\log_{10}{10} ...Law\:\: 5...Log \\[3ex] \log_{10}{10} = 1 ...Law\: 4...Log \\[3ex] \implies \log_{10}{100} = 2 * 1 = 2 \\[3ex] \log_{10}{24} = \log_{10}{2^3 * 3} = \log_{10}{2^3} + \log_{10}{3} ...Law\:\: 1...Log \\[3ex] \log_{10}{2^3} = 3 * \log_{10}{2} ...Law\:\: 5...Log \\[3ex] \implies \log_{10}{2^3} = 3 * 0.3010 = 0.903 \\[3ex] \implies \log_{10}{24} = 0.903 + 0.4771 = 1.3801 \\[3ex] \implies \log_{10}{0.24} = 1.3801 - 2 = -0.6199$
(25.) JAMB If $\log_{10}{2} = 0.3010$ and $\log_{10}{7} = 0.8451$, evaluate $\log_{10}{280}$

We have to use what we were given and what we already know.
Express $280$ in terms of $2's$ and $7's$ and probably $10's$

$280 = 28 * 10 \\[3ex] = 4 * 7 * 10 \\[3ex] = 2^2 * 7 * 10 \\[3ex] \log_{10}{280} = \log_{10}{2^2 * 7 * 10} \\[3ex] = \log_{10}{2^2} + \log_{10}{7} + \log_{10}{10} ...Law\:\: 1...Log \\[3ex] \log_{10}{2^2} = 2 * \log_{10}{2} ...Law\:\: 5...Log \\[3ex] \log_{10}{10} = 1 ...Law\:\: 4...Log \\[3ex] = 2 * 0.3010 + 0.8451 + 1 \\[3ex] = 0.602 + 0.8451 + 1 \\[3ex] = 2.4471$
(26.) ACT What is the value of $\log_2{\sqrt{8}}$?

$\log_2{\sqrt{8}} \\[3ex] \sqrt{8} = 8^{\dfrac{1}{2}} ...Law\:\: 7...Exp \\[3ex] \sqrt{8} = (2^3)^{\dfrac{1}{2}} = 2^{3 * \dfrac{1}{2}} ...Law\:\: 5...Exp \\[3ex] \sqrt{8} = 2^{\dfrac{3}{2}} \\[3ex] = \log_2{2^{\dfrac{3}{2}}} \\[3ex] = \dfrac{3}{2} * \log_2{2} ...Law\:\: 5...Log \\[5ex] \log_2{2} = 1 ...Law\:\: 4...Log \\[3ex] = \dfrac{3}{2} * 1 \\[5ex] = \dfrac{3}{2}$
(27.) WASSCE Without using Tables or Calculator, simplify:

$\dfrac{1}{3}\log{\dfrac{125}{8}} - 2\log{\dfrac{2}{5}} + \log{\dfrac{80}{125}}$

(All logarithms are in base 10)

$\dfrac{1}{3}\log{\dfrac{125}{8}} - 2\log{\dfrac{2}{5}} + \log{\dfrac{80}{125}} \\[7ex] \dfrac{1}{3}\log{\dfrac{125}{8}} = \log{\left(\dfrac{125}{8}\right)}^{\dfrac{1}{3}} ...Law\:\: 5...Log \\[7ex] \left(\dfrac{125}{8}\right)^{\dfrac{1}{3}} = \sqrt[3]{\dfrac{125}{8}} = \dfrac{5}{2} ...Law\:\: 7...Exp \\[5ex] \implies \dfrac{1}{3}\log{\dfrac{125}{8}} = \log{\dfrac{5}{2}} \\[5ex] 2\log{\dfrac{2}{5}} = \log{\dfrac{2}{5}}^2 ...Law\:\: 5...Log \\[5ex] \left(\dfrac{2}{5}\right)^2 = \dfrac{2^2}{5^2} = \dfrac{4}{25} ...Law\:\: 5...Exp \\[5ex] \implies 2\log{\dfrac{2}{5}} = \log{\dfrac{4}{25}} \\[5ex] = \log{\dfrac{5}{2}} - \log{\dfrac{4}{25}} + \log{\dfrac{80}{125}} \\[5ex] = \log{\left(\dfrac{5}{2} \div \dfrac{4}{25} * \dfrac{80}{125}\right)} ...Laws\:\: 1 \:\:and\:\: 2...Log \\[5ex] = \log{\left(\dfrac{5}{2} * \dfrac{25}{4} * \dfrac{80}{125}\right)} \\[5ex] = \log {10} \\[3ex] = 1$
(28.) $e^{\ln{c}}$

$e^{\ln{c}} \\[3ex] = e^{\log_e{c}} \\[3ex] = c ...Law\: 7 ...Log$