I will go over three examples in this indict showing just how to determine algebraically the station of one exponential function. But before you take a look in ~ the operated examples, I indicate that you review the argued steps below very first in order to have actually a great grasp that the general procedure.

You are watching: How to find inverse of exponential function

Steps to find the inverse of an Exponential Function

STEP 1: change f\left( x \right) come y.

\largef\left( x \right) \to y

STEP 2: Interchange \colorbluex and also \colorredy in the equation.

\largex \to y

\largey \to x

STEP 3: isolation the exponential expression top top one side (left or right) the the equation.

The exponential expression shown below is a generic type where b is the base, when N is the exponent.


STEP 4: get rid of the basic b the the exponential expression by taking the logarithms that both sides of the equation.

To do the simplification lot easier, take the logarithm of both sides utilizing the base of the exponential expression itself.Using the log in rule,


STEP 5: fix the exponential equation because that \colorredy to gain the inverse. Finally, replace \colorredy v the train station notation f^ - 1\left( x \right) to compose the final answer.

Replace y with f^ - 1\left( x \right)

Let’s use the argued steps above to solve some problems.

Examples of just how to find the train station of one Exponential Function

Example 1: find the station of the exponential function below.


This have to be basic problem due to the fact that the exponential expression ~ above the right side the the equation is currently isolated for us.

Start by instead of the duty notation f\left( x \right) by y.


Since the exponential expression is by itself on one next of the equation, we have the right to now take it the logarithms the both sides. Once we get the logarithms that both sides, we will use the basic of \colorblue2 due to the fact that this is the basic of the offered exponential expression.

Apply the log in of Exponent dominance which is \log _b\left( b^k \right) = k as component of the simplification process. The dominion states the the logarithm of an exponential number wherein its base is the exact same as the base of the log in is same to the exponent.

We are virtually done! fix for y by including both political parties by 5 then divide the equation by the coefficient the y i m sorry is 3. Don’t forget to change y come f^ - 1\left( x \right). This means that we have found the inverse function.

If us graph the original exponential function and its train station on the exact same XY- plane, they need to be symmetrical along the line \large\colorbluey=x. Which lock are!

The only difference of this problem from the previous one is that the exponential expression has actually a denominator the 2. Various other than that, the actions will it is in the same.

We adjust the duty notation f\left( x \right) come y, adhered to by interchanging the roles of \colorredx and also \colorredy variables.

At this point, us can’t do the action of taking the logarithms of both sides simply yet. The factor is the the exponential expression on the ideal side is not fully by itself. We very first have to get rid of the denominator 2.

We can accomplish that by multiply both political parties of the equation by 2. The left next becomes 2x and also the denominator top top the best side is gone!

By isolating the exponential expression on one side, that is now possible to obtain the logs of both sides. When you execute this, always make sure to use the base of the exponential expression together the basic of the logarithmic operations.

In this case, the basic of the exponential expression is 5. Therefore, we use log operations on both sides making use of thebase of 5.

Using this log in rule, \log _b\left( b^k \right) = k, the fives will certainly cancel the end leaving the exponent \colorblue4x+1 ~ above the ideal side that the equation after ~ simplification. This is great since the log section of the equation is gone.

We have the right to now finish this increase by addressing for the change y, climate replacing that by f^ - 1\left( x \right) to represent that we have derived the train station function.

As you can see, the graphs of the exponential function and that is inverse room symmetrical about the line \large\colorgreeny=x.

I view that we have an exponential expression being split by another. The an excellent thing is that the exponential expressions have actually the same base the 3. We should be able to simplify this making use of the division Rule that Exponent. To division exponential expressions having equal bases, copy the typical base and then subtract their exponents. Below is the rule. The assumption is that b \ne 0.

Observe how the original trouble has been significantly simplified after applying the department Rule the Exponent.

At this point, we have the right to proceed as usual in resolving for the inverse. Rewrite f\left( x \right) as y, adhered to by interchanging the variables \colorredx and also \colorredy.

Before we can get the logs that both sides, isolation the exponential portion of the equation by including both political parties by 4.

Since the exponential expression is using base 3, we take the logs of both sides of the equation with base 3 as well! By law so, the exponent \colorblue2y-1 ~ above the ideal side will certainly drop, so we can continue on solving for y i beg your pardon is the compelled inverse function.

See more: Consecutive Angles In A Parallelogram Are Supplementary, Angles Of A Parallelogram

It verifies the our answer is correct because the graph the the offered exponential functions and also its inverse (logarithmic function) room symmetrical along the line \largey=x.

You might additionally be interested in:

Inverse the a 2×2 Matrix

Inverse the Absolute value Function

Inverse of continuous Function

Inverse of linear Function

Inverse the Logarithmic Function

Inverse the Quadratic Function

Inverse of reasonable Function

Inverse of Square source Function

MATH SUBJECTSIntroductory AlgebraIntermediate AlgebraAdvanced AlgebraAlgebra word ProblemsGeometryIntro come Number TheoryBasic math Proofs
We usage cookies to give you the ideal experience on our website. Please click yes or Scroll under to usage this website with cookies. Otherwise, check your browser settings to revolve cookies turn off or discontinue using the site.OK!Cookie Policy