Question

If f(x)=$5lo{{g}_{5}}$ x $then {{f}^{-1}}(\alpha -\beta)$ where $\alpha$ ,$\beta$ in R is equal to
A. ${{f}^{-1}}(\alpha )-{{f}^{-1}}(\beta )$
B. $\dfrac{{{f}^{-1}}(\alpha )}{{{f}^{-1}}(\beta )}$
C. $\dfrac{1}{f(\alpha -\beta )}$
D. $\dfrac{1}{f(\alpha )-f(\beta )}$

Answer 1

Hint: We have given a function and we are aware that a function that returns the initial value for which a function has produced an output is known as an inverse function. If f(x) is a function that produces the value y, then $f^{-1}(y)$, the inverse function of y, will provide the value x. So here we will first change $f(x) = y$ and obtain x in the terms of y. Do simplify. To get the inverse of the function, replace y with $f^{-1}(x)$ and x with \alpha -\beta.


Formula used:Exponents should be multiplied when raising an exponential equation to a power.
$Log_b M^p = P log_b M$


Complete step by step solution:Given $f(x)=5lo{{g}_{5}}x$
Let $y=5lo{{g}_{5}}x$
In order to determine the value for $f^{-1}(x)$ express x in the terms of y we get;
$\Rightarrow \dfrac{y}{5}=lo{{g}_{5}}x$
$\Rightarrow (x)={{5}^{y/5}}$
Thus,
\Rightarrow ${{f}^{-1}}(x)={{5}^{x/5}}$
Now, for ${{f}^{-1}}(\alpha -\beta )$
$\Rightarrow f^{-1}(\alpha-\beta)=5^{\alpha-\beta /5}$
$\Rightarrow f^{-1}(\alpha-\beta)=5^{\alpha /5}.5^{-\beta /5}$
$\Rightarrow f^{-1}(\alpha-\beta)=\dfrac{{{5}^{\alpha /5}}}{{{5}^{\beta /5}}}$
Hence, ${{f}^{-1}}(\alpha -\beta ) =\dfrac{{{f}^{-1}}(\alpha )}{{{f}^{-1}}(\beta )}$


Thus, Option (B) is correct.

Note: We know a function that can change into another function is referred to as being inverse of the function. In other words, the inverse of a function "f" will take y to x if any function "f" takes x to y. When a function is written as "f" or "F," its inverse is written as "$f^{-1}$" or "$F^{-1}$." Here, (-1) should not be confused with an exponent or a reciprocal.