Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Review / Exponential and Logarithm Equations
Show General Notice Show Mobile Notice Show All Notes Hide All Notes
General Notice

I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.

Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.

Paul
February 18, 2026

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 viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 1.9 : Exponential And Logarithm Equations

1. Find all the solutions to \(12 - 4{{\bf{e}}^{7 + 3\,x}} = 7\). If there are no solutions clearly explain why.

Show All Steps Hide All Steps

Start Solution

There isn’t all that much to do here for this equation. First, we need to isolate the exponential on one side by itself with a coefficient of one.

\[ - 4{{\bf{e}}^{7 + 3\,x}} = - 5\hspace{0.5in} \Rightarrow \hspace{0.5in}{{\bf{e}}^{7 + 3\,x}} = \frac{5}{4}\] Show Step 2

Now all we need to do is take the natural logarithm of both sides and then solve for \(x\).

\[\begin{align*}\ln \left( {{{\bf{e}}^{7 + 3\,x}}} \right) & = \ln \left( {\frac{5}{4}} \right)\\ 7 + 3x & = \ln \left( {\frac{5}{4}} \right)\\ x & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\left( {\ln \left( {\frac{5}{4}} \right) - 7} \right) = - 2.25895}}\end{align*}\]

Depending upon your preferences either the exact or decimal solution can be used.