Question

How do you find the inverse of $y={{4}^{x}}$?

Answer 1

Any function can have its inverse function, if it is a bijective function. A bijective function is both one-one and onto function. That is, each image of the function has distinct preimage and range equals to its co-domain. We can write the inverse function equation of an invertible function by expressing x in terms of y and then replacing with x.

Complete step by step answer:
As per the question, we are asked to find out the inverse function of the function $y={{4}^{x}}$. We know that the graph of $y={{4}^{x}}$ function increases with x. So, we can say that it is an increasing function in its domain. As x value ranges from $-\infty$ to $\infty$, y value ranges from 0 to $\infty$. Hence, $y={{4}^{x}}$ is an invertible function.
Now, let us find the inverse function of $y={{4}^{x}}$.
Given equation is,
$\Rightarrow y={{4}^{x}}$
Here, we have 4 to the power x. So, we have to apply natural logarithm on both sides of the given equation. Then we get,
$\Rightarrow \ln y=\ln {{4}^{x}}$ -------(1)
We know that, if we have the logarithm of a number ‘a’ to the power ‘b’, we have to write ‘b’ as the coefficient of the logarithm of ‘a’. That is, we can write
$\Rightarrow \ln {{4}^{x}}=x\ln 4$ ------(2)
By substituting the equation (2) into the equation (1), we get
$\Rightarrow \ln y=x\ln 4$
$\therefore x=\dfrac{\ln y}{\ln 4}$ --------(3)
We know that, $\ln 4=2\ln 2=2\times 0.693$. That is $\ln 4=1.386$ .
$\Rightarrow y=0.7213\ln x$
By interchanging x and y we got the above equation.

$\therefore y=0.7213\ln x$ is the required inverse function equation of $y={{4}^{x}}$.

Note: We need to verify the inverse function obtained by composing $f$ and ${{f}^{-1}}$. Common errors while composing functions: Students sometimes forget where each of the functions is defined before composing functions, which lead to non – existing results. They also sometimes forget that composition is not a commutative operation, that is, $f\circ g\ne g\circ f$. Also, the graphs of $f$ and ${{f}^{-1}}$ are symmetric about the line y=x.