Question

If $\dfrac{{{\log }_{10}}25}{{{\log }_{10}}5}={{\log }_{10}}x$, then the value of x is equal to
(a) 10
(b) 5
(c) 2
(d) 100

Answer 1

Hint: We start solving the problem by using the fact $\dfrac{{{\log }_{a}}b}{{{\log }_{a}}c}={{\log }_{c}}b$ for the left-hand side of the given equation. We then use the fact ${{\log }_{a}}{{b}^{n}}=n{{\log }_{a}}b$ to proceed through the problem. We then use ${{\log }_{a}}a=1$ to proceed further through the problem. We then use the fact that if ${{\log }_{a}}b=c$, then $b={{a}^{c}}$ and make necessary calculations to get the desired value of x.

Complete step-by-step answer:
According to the problem, we have the value of $\dfrac{{{\log }_{10}}25}{{{\log }_{10}}5}$ equals ${{\log }_{10}}x$. We need to find the value of x.
We have $\dfrac{{{\log }_{10}}25}{{{\log }_{10}}5}={{\log }_{10}}x$ ---(1).
We know that $\dfrac{{{\log }_{a}}b}{{{\log }_{a}}c}={{\log }_{c}}b$ for positive values of ‘a’ (except1), ‘b’ and ‘c’ (except1). We use this result in equation (1).
$\Rightarrow {{\log }_{5}}25={{\log }_{10}}x$.
$\Rightarrow {{\log }_{5}}{{5}^{2}}={{\log }_{10}}x$ ---(2).
We know that ${{\log }_{a}}{{b}^{n}}=n{{\log }_{a}}b$for positive values of ‘a’ (except 1) and ‘b’. We use this result in equation (2).
$\Rightarrow 2{{\log }_{5}}5={{\log }_{10}}x$ ---(3).
We know that ${{\log }_{a}}a=1$ for any positive value of ‘a’ (except 1). We use this result in equation (3).
$\Rightarrow 2\times 1={{\log }_{10}}x$.
$\Rightarrow 2={{\log }_{10}}x$ ---(4).
We know that if ${{\log }_{a}}b=c$, then $b={{a}^{c}}$ for any positive values of ‘a’ (except 1) and ‘b’. We use this result in equation (4).
$\Rightarrow x={{10}^{2}}$.
$\Rightarrow x=100$.
We have found the value of x as 100.
∴ The value of x is 100.
The correct option for the given problem is (d).

Note: Before solving the problem, we need to make sure that the base and number that we are applying log are positive. Because logarithms apply only for the positive base (except 1). We should not directly divide the numbers that were not present in base i.e., $\dfrac{{{\log }_{a}}x}{{{\log }_{a}}y}\ne {{\log }_{a}}\left( \dfrac{x}{y} \right)$. Similarly, we can expect problems that require use of multiplication and division properties of logarithms.