Paul's Online Notes
Home / Calculus I / Review / Exponential and Logarithm 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 1-9 : Exponential And Logarithm Equations

2. Find all the solutions to $$1 = 10 - 3{{\bf{e}}^{{z^{\,2}} - 2\,z}}$$. 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.

$- 9 = - 3{{\bf{e}}^{{z^{\,2}} - 2\,z}}\hspace{0.5in} \Rightarrow \hspace{0.5in}{{\bf{e}}^{{z^{\,2}} - 2\,z}} = 3$ Show Step 2

Now all we need to do is take the natural logarithm of both sides and then solve for $$z$$.

\begin{align*}\ln \left( {{{\bf{e}}^{{z^{\,2}} - 2\,z}}} \right) & = \ln \left( 3 \right)\\ {z^2} - 2z & = \ln \left( 3 \right)\\ {z^2} - 2z - \ln \left( 3 \right) & = 0\end{align*}

Now, before proceeding with the solution here let’s pause and make sure that we don’t get too excited about the “strangeness” of the quadratic above. If we’d had the quadratic,

${z^2} - 2z - 5 = 0$

for instance, we’d know that all we would need to do is use the quadratic formula to get the solutions.

That’s all we need to as well for the quadratic that we have from our work. Of course we don’t have a 5 we have a ln(3), but ln(3) is just a number and so we can use the quadratic formula to find the solutions here as well. Here is the work for that.

\begin{align*}z & = \frac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - \ln \left( 3 \right)} \right)} }}{{2\left( 1 \right)}} = \frac{{2 \pm \sqrt {4 + 4\ln \left( 3 \right)} }}{2}\\ & = \frac{{2 \pm \sqrt {4\left( {1 + \ln \left( 3 \right)} \right)} }}{2} = \frac{{2 \pm 2\sqrt {1 + \ln \left( 3 \right)} }}{2} = \require{bbox} \bbox[2pt,border:1px solid black]{{1 \pm \sqrt {1 + \ln \left( 3 \right)} = - 0.4487,\,\,\,2.4487}}\end{align*}

Notice that we did a little simplification on the root. This doesn’t need to be done, but can make the exact solution a little easier to deal with. Also, depending upon your preferences either the exact or decimal solution can be used.

Before leaving this solution we should again make a point that not all quadratics will be the “simple” type of quadratics that you may be used to solving from an Algebra class. They can, and often will be, messier that those. That doesn’t mean that you can’t solve them. They are, for all intents and purposes, identical to the types of problems you are used to working. The only real difference is that they numbers are a little messier.

So, don’t get too excited about this kind of problem. They will happen on occasion and you are capable of solving them!