Question

The value of ${\log _{10}}75 + {\log _{10}}28 - {\log _{10}}21$ is equal to
A.${\log _{10}}81$
B.${\log _{10}}20$
C.2
D.None of these

Answer 1

Hint: Simplify the first two terms of the given expression using the formula ${\log _a}b + {\log _a}c = {\log _a}\left( {b \times c} \right)$ and solve the resultant and the last term using the formula, ${\log _a}b - {\log _a}c = {\log _a}\left( {\dfrac{b}{a}} \right)$. Then get the resultant answer using the formula of log, ${\log _a}{m^2} = 2{\log _a}m$ and ${\log _a}a = 1$

Complete step-by-step answer:
Let us begin the question by simplifying the first two terms of the given expression.
The first log is added. Therefore, use the formula, ${\log _a}b + {\log _a}c = {\log _a}\left( {b \times c} \right)$
Hence, the first two terms in the expression ${\log _{10}}75 + {\log _{10}}28 - {\log _{10}}21$ can be simplified as ${\log _{10}}75 + {\log _{10}}28 = {\log _{10}}\left( {75 \times 28} \right) \\\ {\log _{10}}75 + {\log _{10}}28 = {\log _{10}}\left( {2100} \right) \\\$
Now, the given expression can be written as, ${\log _{10}}2100 - {\log _{10}}21$
We can simplify the above expression using the formula, ${\log _a}b - {\log _a}c = {\log _a}\left( {\dfrac{b}{a}} \right)$
The equation can be written as,
${\log _{10}}2100 - {\log _{10}}21 = {\log _{10}}\dfrac{{2100}}{{21}} \\\ {\log _{10}}2100 - {\log _{10}}21 = {\log _{10}}100 \\\ {\log _{10}}2100 - {\log _{10}}21 = {\log _{10}}{\left( {10} \right)^2} \\\$
By using the identity, ${\log _a}{m^2} = 2{\log _a}m$, we can rewrite ${\log _{10}}{\left( {10} \right)^2}$ this as $2{\log _{10}}10$
Also, ${\log _a}a = 1$, therefore, $2{\log _{10}}10 = 2\left( 1 \right) = 2$
Hence, the value of ${\log _{10}}75 + {\log _{10}}28 - {\log _{10}}21$ is equal to 2.
Hence, option C is the correct one.

Note: In this question identities of logarithm are used to solve it, such as${\log _a}b + {\log _a}c = {\log _a}\left( {b \times c} \right)$, ${\log _a}b - {\log _a}c = {\log _a}\left( {\dfrac{b}{a}} \right)$ , ${\log _a}{m^2} = 2{\log _a}m$ and ${\log _a}a = 1$. The identities are used to simplify the expression.