Question

If $\log 4 + 2\log 3$ is $\log m$ ,then the value of $m$ is equal to
(a) $33$
(b) $31$
(c) $36$
(d) $38$

Answer 1

Hint : For solving this logarithmic related problem, we will generally use the properties of logarithmic functions such as ${\log _a}m \times n = {\log _a}m + {\log _a}n$ and ${\log _a}{m^s} = s{\log _a}m$ .To solve this question we will first consider left-hand side of the expression i.e., $\log 4 + 2\log 3$ and arrange the logarithmic terms to the simplest form using these standard results and then equalize it with the right hand side of the expression i.e., $\log m$ and simplify the expression to find the value of $m$

Complete step-by-step answer :
Now, in the question we are given that,
$\log 4 + 2\log 3$ is $\log m$
$\Rightarrow \log 4 + 2\log 3 = \log m{\text{ }} - - - \left( 1 \right)$
Now, consider left-hand side i.e.,
$\log 4 + 2\log 3{\text{ }} - - - \left( 2 \right)$
Now, we know that property of logarithmic function which is
${\log _a}{m^s} = s{\log _a}m$
So, we can write $2\log 3$ as $\log {3^2}$
Therefore, equation $\left( 2 \right)$ becomes,
$\log 4 + \log {3^2}$
which can also be written as
$\log 4 + \log 9$
Now, we know that the property of logarithmic function which is
${\log _a}m \times n = {\log _a}m + {\log _a}n$
Therefore, above expression can be written as,
$\log \left( {4 \times 9} \right)$
$= \log \left( {36} \right)$
Therefore, we get
$\Rightarrow \log 4 + 2\log 3 = \log \left( {36} \right){\text{ }} - - - \left( 3 \right)$
Now, on comparing equation $\left( 3 \right)$ with equation $\left( 1 \right)$ we get,
$\log m = \log 36$
We know that,
If $\log x = \log y$ then $x = y$
Therefore, we get
$m = 36$
Thus, if $\log 4 + 2\log 3$ is $\log m$ then the value of $m$ is equal to $36$
So, the correct answer is “Option C”.

Note : The important condition while applying the laws of the logarithm is that the base of the logarithm functions involved should be the same in all the calculations. If the base of the logarithm is not given, we can consider the base as $10$ i.e., common logarithm. Also, one of the most basic tricks to solve logarithmic functions is to rearrange the order of base value and variable value such that you are left with some expression to which you can replace with some standard results.
Some other useful formulas are:
${\log _a}\dfrac{m}{n} = {\log _a}m - {\log _a}n$
${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$
${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$
${\log _a}{a^a} = a$