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### Section 6-3 : Solving Exponential Equations

8. Solve the following equation.

\[{{\bf{e}}^{7 + 2x}} - 3 = 0\]Show All Steps Hide All Steps

Start SolutionBefore we put any logarithms into this problem we first need to get the exponential on one side by itself so let’s do that first.

\[{{\bf{e}}^{7 + 2x}} = 3\] Show Step 2Now we can take the logarithm of both sides and because we have a base of \(\bf{e}\) in this problem the natural logarithm is probably the best choice. So, taking the logarithm (using the natural logarithm) of both sides gives,

\[\ln {{\bf{e}}^{7 + 2x}} = \ln 3\] Show Step 3Now we can use the logarithm property that says,

\[\ln {{\bf{e}}^{f\left( x \right)}} = f\left( x \right)\]to simplify the left side of the equation. Doing this gives,

\[7 + 2x = \ln 3\] Show Step 4Finally, 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.

\[\begin{align*}7 + 2x & = \ln 3\\ 2x & = \ln 3 - 7\\ x & = \frac{{\ln 3 - 7}}{2} = \frac{{1.098612289 - 7}}{2} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 2.950693856}}\end{align*}\]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.

Also, be careful when evaluating the numerator in the final answer. The 7 was outside of the logarithm and so cannot be moved into the logarithm. We probably should have been a little more careful with parenthesis and written the answer as,

\[x = \frac{{\ln \left( 3 \right) - 7}}{2}\]which makes it a little clearer that the 7 isn’t inside the logarithm. However, we typically don’t put the parenthesis on the logarithm when it is just a number.