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

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.