Finding the inverse of a log function is as easy as following the suggested steps below. You will realize later after seeing some examples that most of the work boils down to solving an equation. The key steps involved include isolating the log expression and then rewriting the log equation into an exponential equation. You will see what I mean when you go over the worked examples below.
STEP 1: Replace the function notation [latex]f\left( x \right)[/latex] by [latex]y[/latex].
[latex]f\left( x \right) \to y[/latex]
STEP 2: Switch the roles of [latex]x[/latex] and [latex]y[/latex].
[latex]x \to y[/latex]
[latex]y \to x[/latex]
STEP 3: Isolate the log expression on one side (left or right) of the equation.
STEP 4: Convert or transform the log equation into its equivalent exponential equation.
STEP 5: Solve the exponential equation for [latex]y[/latex] to get the inverse. Then replace [latex]y[/latex] by [latex]>\left( x \right)[/latex] which is the inverse notation to write the final answer.
Rewrite [latex]\colory[/latex] as [latex]\color>\left( x \right)[/latex]
Example 1: Find the inverse of the log equation below.
[latex]f\left( x \right) = <\log _2>\left( \right)[/latex]
Start by replacing the function notation [latex]f\left( x \right)[/latex] by [latex]y[/latex]. Then, interchange the roles of [latex]\colorx[/latex] and [latex]\colory[/latex].
Proceed by solving for [latex]y[/latex] and replacing it by [latex]>\left( x \right)[/latex] to get the inverse. Part of the solution below includes rewriting the log equation into an exponential equation. Here’s the formula again that is used in the conversion process.
Notice how the base [latex]2[/latex] of the log expression becomes the base with an exponent of [latex]x[/latex]. The stuff inside the parenthesis remains in its original location.
Once the log expression is gone by converting it into an exponential expression, we can finish this off by subtracting both sides by [latex]3[/latex]. Don’t forget to replace the variable y by the inverse notation [latex]>\left( x \right)[/latex] the end.
One way to check if we got the correct inverse is to graph both the log equation and inverse function in a single [latex]xy[/latex]-axis. If their graphs are symmetrical along the line [latex]\largey = x>[/latex], then we can be confident that our answer is indeed correct.
Example 2: Find the inverse of the log function
[latex]f\left( x \right) = <\log _5>\left( \right) – 7[/latex]
Let’s add up some level of difficulty to this problem. The equation has a log expression being subtracted by [latex]7[/latex]. I hope you can assess that this problem is extremely doable. The solution will be a bit messy but definitely manageable.
So I begin by changing the [latex]f\left( x \right)[/latex] into [latex]y[/latex], and swapping the roles of [latex]\colorx[/latex] and [latex]\colory[/latex].
Now, we can solve for [latex]y[/latex]. Add both sides of the equation by [latex]7[/latex] to isolate the logarithmic expression on the right side.
By successfully isolating the log expression on the right, we are ready to convert this into an exponential equation. Observe that the base of log expression which is [latex]6[/latex] becomes the base of the exponential expression on the left side. The expression [latex]2y-1[/latex] inside the parenthesis on the right is now by itself without the log operation.
After doing so, proceed by solving for [latex]\colory[/latex] to obtain the required inverse function. Do that by adding both sides by [latex]1[/latex], followed by dividing both sides by the coefficient of [latex]\colory[/latex] which is [latex]2[/latex].
![<[5^(x+7)]+1></p>
<p>/2 = (f^-1)(x)](https://www.chilimath.com/wp-content/uploads/2017/02/ex2_s3-1.png)
Let’s sketch the graphs of the log and inverse functions in the same Cartesian plane to verify that they are indeed symmetrical along the line [latex]\largey=x>[/latex].
Example 3: Find the inverse of the log function
So this is a little more interesting than the first two problems. Observe that the base of log expression is missing. If you encounter something like this, the assumption is that we are working with a logarithmic expression with base [latex]10[/latex]. Always remember this concept to help you get around problems with the same setup.
I hope you are already more comfortable with the procedures. We start again by making [latex]f\left( x \right)[/latex] as [latex]y[/latex], then switching around the variables [latex]\colorx[/latex] and [latex]\colory[/latex] in the equation.
Our next goal is to isolate the log expression. We can do that by subtracting both sides by [latex]1[/latex] followed by dividing both sides by [latex]-3[/latex].
The log expression is now by itself. Remember, the “missing” base in the log expression implies a base of [latex]10[/latex]. Transform this into an exponential equation, and start solving for [latex]y[/latex].
Notice that the entire expression on the left side of the equation becomes the exponent of [latex]10[/latex] which is the implied base as pointed out before.
Continue solving for [latex]y[/latex] by subtracting both sides by [latex]1[/latex] and dividing by [latex]-4[/latex]. After [latex]y[/latex] is fully isolated, replace that by the inverse notation [latex]\large>\left( x \right)>[/latex]. Done!
![[(1-10)^(1-x/3)]/4 = (f^-1)(x)](https://www.chilimath.com/wp-content/uploads/2017/02/ex3_s3-1.png)
Graphing the original function and its inverse on the same [latex]xy[/latex]-axis reveals that they are symmetrical about the line [latex]\largey=x>[/latex].
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