Question

If \[{\text{log 15 = a}}\] and \[{\text{log 75 = b}}\] then \[{\text{log7545}}\] is:
A) \[\dfrac{{3b - a}}{a}\]
B) \[\dfrac{{b - 3a}}{a}\]
C) \[\dfrac{{3a - b}}{b}\]
D) \[\dfrac{{a - 3b}}{b}\]

Answer 1

Hint: In order to solve this problem we need to convert each of the logs to the base of $e$ and then apply the property of the logarithm function to find the desired value.
Properties of log function are:
\[\log {\text{ab = }}\dfrac{{\log eb}}{{\log ea}}\]
\[{\text{log_e}}\left( {{\text{ab}}} \right){\text{ = log_ea + log_eb}}\]
\[{\text{log_e}}{{\text{a}}^{\text{b}}} = {\text{blog_ea}}\]

Complete step by step solution:
We are given \[{\text{log 15 = a}}\] and \[{\text{log 75 = b}}\] and we need to find the value of \[{\text{log7545}}\].
First we will convert the \[{\text{log7545}}\] to the log with base e by using the following property of log functions:
\[\log {\text{ab = }}\dfrac{{\log eb}}{{\log ea}}\] (property 1)
Hence by applying the property 1 we get:
\[{\text{log7545}} = \dfrac{{\log e45}}{{\log e75}}..............(1)\]
Also, we know that
\[
 {\text{log_e 15}} = {\text{log_e }}\left( {3 \times 5} \right) \\
 {\text{log_e 75}} = {\text{log_e }}\left( {3 \times {5^2}} \right) \\
 \log e45 = {\text{log_e }}\left( {{3^2} \times 5} \right) \\
 \]
Now using another property of log functions for above values:
\[{\text{log_e}}\left( {{\text{ab}}} \right){\text{ = log_ea + log_eb}}\] (property 2)
Applying the property 2 on \[{\text{log_e 15}}\] we get:
\[
 {\text{log_e 15}} = {\text{log_e3 + log_e5}} \\
 {\text{a}} = {\text{log_e3 + log_e5 }}...............\left( 2 \right) \\
 \]
Now applying the property 2 on \[{\text{log_e 75}}\] we get:
\[{\text{log_e 75}} = {\text{log_e3 + log_e}}{{\text{5}}^2}\]
We can use another property of log functions here:
\[{\text{log_e}}{{\text{a}}^{\text{b}}} = {\text{blog_ea}}\] (property 3)
Applying property 3 we get:
\[
 {\text{log_e 75}} = {\text{log_e3 + 2log_e5}} \\
 {\text{b = log_e3 + 2log_e5 }}................\left( 3 \right) \\
 \]
Now, applying property 2 on \[\log e45\] we get:
\[\log e45 = {\text{log_e}}{{\text{3}}^2}{\text{ + log_e5}}\]
Applying property 3 now on this equation we get:
\[\log e45 = 2{\text{log_e3 + log_e5 }}..............\left( 4 \right)\]
Now solving the equation 2 and equation 3 by elimination method to get values of \[{\text{log_e3}}\] and \[{\text{log_e5}}\] in terms of a and b :
Multiplying equation 2 by 2 and then subtracting equation 3 from equation 2 we get:
\[
 {\text{2a}} - {\text{b}} = 2{\text{log_e3 + 2log_e5}} - {\text{log_e3}} - {\text{2log_e5}} \\
 {\text{2a}} - {\text{b = log_e3}} \\
 {\text{b}} - {\text{a = log_e5}} \\
 \]
Now putting these values in equation 4 we get:
\[
 \log e45 = 2\left( {{\text{2a}} - {\text{b}}} \right){\text{ + b}} - {\text{a}} \\
 \log e45 = 4{\text{a}} - {\text{2b}} + {\text{b}} - {\text{a}} \\
 \log e45 = 3{\text{a}} - {\text{b}} \\
 \]
Now putting the values of \[\log e45\] and \[{\text{log_e 75}}\] in equation 1 we get:
\[{\text{log7545}} = \dfrac{{{\text{3a}} - {\text{b}}}}{{\text{b}}}\]

Option (C) is correct.

Note:
Students should keep in mind that the quantity inside the logarithm function can never be zero as the logarithm function is not defined at zero.
Also, the logarithm function is a strictly increasing function.