Question

How do you find the inverse of $y = {\log _{\dfrac{1}{4}}}x$?

Answer 1

This problem deals with solving the given equation in logarithms. Here some basic and fundamental properties or identities of logarithms are used. Along with some concepts of logarithms, some basic concepts or identities from exponents and bases are used, such as given below:
If ${\log _a}b = x$, then $b = {a^x}$.
$\Rightarrow {\left( {{a^m}} \right)^n} = {\left( a \right)^{mn}}$

Complete step-by-step answer:
Consider the given equation as shown below:
$\Rightarrow y = {\log _{\dfrac{1}{4}}}x$
Here the given equation is a function of the variable of $x$, so here $y = f\left( x \right)$, as shown below:
$\Rightarrow f\left( x \right) = {\log _{\dfrac{1}{4}}}x$
The inverse of the function $y = f\left( x \right)$ is $x = f\left( y \right)$.
Now we have to find the inverse function $f\left( x \right)$ which is $f\left( y \right)$, where the function $f\left( y \right)$ is a function varying in $x$.
$\Rightarrow y = {\log _{\dfrac{1}{4}}}x$
Applying the basic and fundamental property of logarithms which is, If ${\log _a}b = x$ then $b = {a^x}$, to the above equation as shown below:
$\Rightarrow x = {\left( {\dfrac{1}{4}} \right)^y}$
Now simplifying the right hand side of the above equation as shown below:
We know that $4 = {2^2}$, then the inverse of $4$ which is equal to $\dfrac{1}{4} = \dfrac{1}{{{2^2}}}$, so substituting this value in the above equation as shown below:
$\Rightarrow x = {\left( {\dfrac{1}{{{2^2}}}} \right)^y}$
We know that $\dfrac{1}{{{2^2}}} = {2^{ - 2}}$, substituting this in the above equation as shown below:
$\Rightarrow x = {\left( {{2^{ - 2}}} \right)^y}$
Now applying one of the property from exponents and bases which is ${\left( {{a^m}} \right)^n} = {\left( a \right)^{mn}}$ to the above equation as shown below:
$\Rightarrow x = {2^{ - 2y}}$
Now we obtained that $x = f\left( y \right)$, so the inverse function is given by $f\left( y \right) = {f^{ - 1}}\left( x \right) = x$
$\Rightarrow {f^{ - 1}}\left( x \right) = {2^{ - 2y}}$
Final answer: The inverse of $y = {\log _{\dfrac{1}{4}}}x$ is $x = {2^{ - 2y}}$.
Note: Please note that to solve this problem, few of the basic identities or properties of logarithms and exponents and bases are used and applied in the above problem. Also there are few more identities which are given by:
$\Rightarrow \log ab = \log a + \log b$
$\Rightarrow \log \left( {\dfrac{a}{b}} \right) = \log a - \log b$
$\Rightarrow {e^{{{\log }_e}a}} = a$