Question

If ${{\text{a}}^{\text{x}}}{\text{ = }}{{\text{b}}^{\text{y}}}$, then
A. ${\text{log}}\dfrac{{\text{a}}}{{\text{b}}}{\text{ = }}\dfrac{{\text{x}}}{{\text{y}}}$
B. $\dfrac{{{\text{loga}}}}{{{\text{logb}}}}{\text{ = }}\dfrac{{\text{x}}}{{\text{y}}}$
C. $\dfrac{{{\text{loga}}}}{{{\text{logb}}}}{\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}$
D. None of these

Answer 1

Hint: To solve this question we need to know the basic theory related to the logarithm. As we know logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers. Here we use logarithm properties to solve ${{\text{a}}^{\text{x}}}{\text{ = }}{{\text{b}}^{\text{y}}}$.

Complete step-by-step answer:
We have,
${{\text{a}}^{\text{x}}}{\text{ = }}{{\text{b}}^{\text{y}}}$
Taking logarithm both side
${\text{log(a}}{{\text{)}}^{\text{x}}}{\text{ = log(b}}{{\text{)}}^{\text{y}}}$
As we know, the exponential rule of logarithms which states that the logarithm of m with a rational exponent is equal to the exponent times its logarithm
                                                   ${\text{log}}{\left( {\text{m}} \right)^{\text{n}}}{\text{ = n log(m)}}$
$ \Rightarrow $${\text{x log(a) = y log(b)}}$
$ \Rightarrow $ ${\text{log(a) = }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ log(b)}}$
$ \Rightarrow $ $\dfrac{{{\text{log(a)}}}}{{{\text{log(b)}}}}{\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ }}$
Therefore, If ${{\text{a}}^{\text{x}}}{\text{ = }}{{\text{b}}^{\text{y}}}$, then $\dfrac{{{\text{log(a)}}}}{{{\text{log(b)}}}}{\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}{\text{ }}$.
Thus, option (C) is the correct answer.

Note: A logarithm has various important properties that prove multiplication and division of logarithms can also be written in the form of logarithm of addition and subtraction. i.e. ${{\text{b}}^{\text{y}}}{\text{ = a}}$ and it is read as “the logarithm of a to base b.”