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Solving Exponential Equations

In each of the equations in this section the problem is there is a variable in the exponent. In order to solve these, we will need to get the variable out of the exponent. This means using Property 3 and/or 7 from the Logarithm Properties section above. In most cases it will be easier to use Property 3 if possible. So, pick an appropriate logarithm and take the log of both sides, then use Property 3 (or Property 7) where appropriate to simplify. Note that often some simplification will need to be done before taking the logs.

Solve each of the following equations. Show All Solutions Hide All Solutions

  1. \(2{{\bf{e}}^{4x - 2}} = 9\)
    Show Solution

    The first thing to note is that Property 3 is \({\log _b}{b^x} = x\) and NOT \({\log _b}\left( {2{b^x}} \right) = x\)! In other words, we’ve got to isolate the exponential on one side by itself with a coefficient of 1 (one) before we take logs of both sides.

    We’ll also need to pick an appropriate log to use. In this case the natural log would be best since the exponential in the problem is \({{\bf{e}}^{4x - 2}}\). So, first isolate the exponential on one side.

    \[\begin{align*}2{{\bf{e}}^{4x - 2}} & = 9\\ {{\bf{e}}^{4x - 2}} & = \frac{9}{2}\end{align*}\]

    Now, take the natural log of both sides and use Property 3 to simplify.

    \[\begin{align*}\ln \left( {{{\bf{e}}^{4x - 2}}} \right) & = \ln \left( {\frac{9}{2}} \right)\\ 4x - 2 & = \ln \left( {\frac{9}{2}} \right)\end{align*}\]

    Now you all can solve \(4x - 2 = 9\) so you can solve the equation above. All you need to remember is that \(\ln \left( {\frac{9}{2}} \right)\) is just a number, just as 9 is a number. So, add 2 to both sides, then divide by 4 (or multiply by \(\frac{1}{4}\)).

    \[\begin{align*}4x & = 2 + \ln \left( {\frac{9}{2}} \right)\\ x & = \frac{1}{4}\left( {2 + \ln \left( {\frac{9}{2}} \right)} \right)\\ x & = 0.8760193492\end{align*}\]

    Note that while the natural logarithm was the easiest (since the left side simplified down nicely) we could have used any other log had we wanted to. For instance, we could have used the common log as follows. Remember that in this case we won’t be able to use Property 3 as this requires both the log and the exponential to have the same base which won’t be the case here. Therefore, we’ll need to use Property 7 to do the simplification.

    \[\begin{align*}\log \left( {{{\bf{e}}^{4x - 2}}} \right) & = \log \left( {\frac{9}{2}} \right)\\ \left( {4x - 2} \right)\log {\bf{e}} & = \log \left( {\frac{9}{2}} \right)\end{align*}\]

    As you can see the problem here is that we’ve got a \(\log {\bf{e}}\) left over after using Property 7. While this can be dealt with using a calculator, it adds a complexity to the problem that should be avoided if at all possible. Solving gives us

    \[\begin{align*}4x - 2 & = \frac{{\log \left( {\frac{9}{2}} \right)}}{{\log {\bf{e}}}}\\ 4x & = 2 + \frac{{\log \left( {\frac{9}{2}} \right)}}{{\log {\bf{e}}}}\\ x & = \frac{1}{4}\left( {2 + \frac{{\log \left( {\frac{9}{2}} \right)}}{{\log {\bf{e}}}}} \right)\\ x & = 0.8760193492\end{align*}\]
  2. \({10^{{t^2} - t}} = 100\)
    Show Solution

    Now, in this case it looks like the best logarithm to use is the common logarithm since left hand side has a base of 10. There’s no initial simplification to do, so just take the log of both sides and simplify.

    \[\begin{align*}\log {10^{{t^2} - t}} & = \log 100\\ {t^2} - t & = 2\end{align*}\]

    At this point, we’ve just got a quadratic that can be solved

    \[\begin{align*}{t^2} - t - 2 & = 0\\ \left( {t - 2} \right)\left( {t + 1} \right) & = 0\end{align*}\]

    So, it looks like the solutions in this case are \(t = 2\) and \(t = - 1\).

    As with the last one you could use a different log here, but it would have made the quadratic significantly messier to solve.

  3. \(7 + 15{{\bf{e}}^{1 - 3z}} = 10\)
    Show Solution

    There’s a little more initial simplification to do here, but other than that it’s similar to the first problem in this section.

    \[\begin{align*}7 + 15{{\bf{e}}^{1 - 3z}} & = 10\\ 15{{\bf{e}}^{1 - 3z}} & = 3\\ {{\bf{e}}^{1 - 3z}} & = \frac{1}{5}\end{align*}\]

    Now, take the log and solve. Again, we’ll use the natural logarithm here.

    \[\begin{align*}\ln \left( {{{\bf{e}}^{1 - 3z}}} \right) & = \ln \left( {\frac{1}{5}} \right)\\ 1 - 3z & = \ln \left( {\frac{1}{5}} \right)\\ - 3z & = - 1 + \ln \left( {\frac{1}{5}} \right)\\ z & = - \frac{1}{3}\left( { - 1 + \ln \left( {\frac{1}{5}} \right)} \right)\\ z & = 0.{\rm{8698126372}}\end{align*}\]
  4. \(x - x{{\bf{e}}^{5x + 2}} = 0\)
    Show Solution

    This one is a little different from the previous problems in this section since it’s got \(x\)’s both in the exponent and out of the exponent. The first step is to factor an \(x\) out of both terms.

    NOT DIVIDE AN \(x\) FROM BOTH TERMS!!!!

    \[\begin{align*}x - x{{\bf{e}}^{5x + 2}} & = 0\\ x\left( {1 - {{\bf{e}}^{5x + 2}}} \right) & = 0\end{align*}\]

    So, it’s now a little easier to deal with. From this we can see that we get one of two possibilities.

    \[\begin{align*}x & = 0\hspace{0.25in}{\rm{OR}}\\ 1 - {{\bf{e}}^{5x + 2}} & = 0\end{align*}\]

    The first possibility has nothing more to do, except notice that if we had divided both sides by an \(x\) we would have missed this one so be careful. In the second possibility we’ve got a little more to do. This is an equation similar to the first few that we did in this section.

    \[\begin{align*}{{\bf{e}}^{5x + 2}} & = 1\\ 5x + 2 & = \ln 1\\ 5x + 2 & = 0\\ x & = - \frac{2}{5}\end{align*}\]

    Don’t forget that \(\ln 1 = 0\)!

    So, the two solutions are \(x = 0\) and \(x = - \frac{2}{5}\).