Paul's Online Notes
Home / Algebra / Exponential and Logarithm Functions / Solving Exponential Equations
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 6-3 : Solving Exponential Equations

6. Solve the following equation.

${7^{1 - x}} = {4^{3x + 1}}$

Show All Steps Hide All Steps

Start Solution

For this equation there is no way to easily get both sides with the same base. Therefore, we’ll need to take the logarithm of both sides.

We can use any logarithm and the natural logarithm and common logarithm are usually good choices since most calculators can handle them. In this case there really isn’t any reason to use one or the other so we’ll use the natural logarithm (it’s easier to write two letters – ln versus three letters – log after all…).

Taking the logarithm (using the natural logarithm) of both sides gives,

$\ln {7^{1 - x}} = \ln {4^{3x + 1}}$ Show Step 2

Now we can use the logarithm property that says,

${\log _b}{x^r} = r{\log _b}x$

to move the exponents out of each of the logarithms. Doing this gives,

$\left( {1 - x} \right)\ln 7 = \left( {3x + 1} \right)\ln 4$ Show Step 3

Finally, all we need to do is solve for $$x$$. Recall that the equations at this step tend to look messier than we are used to dealing with. However, the logarithms in the equations at this point are just numbers and so we treat them as we treat all numbers with these kinds of equations. The work will be messier than we are used to but just keep in mind that the logarithms are just numbers!

Here is the rest of the work for this problem.

\begin{align*}\left( {1 - x} \right)\ln 7 & = \left( {3x + 1} \right)\ln 4\\ \ln 7 - x\ln 7 & = 3x\ln 4 + \ln 4\\ \ln 7 - \ln 4 & = 3x\ln 4 + x\ln 7\\ \ln 7 - \ln 4 & = \left( {3\ln 4 + \ln 7} \right)x\\ x & = \frac{{\ln 7 - \ln 4}}{{3\ln 4 + \ln 7}} = \frac{{1.945910149 - 1.386294361}}{{3\left( {1.386294361} \right) + 1.945910149}} = \require{bbox} \bbox[2pt,border:1px solid black]{{0.091668262}}\end{align*}

Again, the work is messier than we are used to but it is not really different from work we’ve done previously in solving equations. The answer is also going to be “messier” in the sense that it is a decimal and is liable to almost always be a decimal for most of these types of problems so don’t worry about that.