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

10. Find all the solutions to $$2\log \left( z \right) - \log \left( {7z - 1} \right) = 0$$. If there are no solutions clearly explain why.

Show All Steps Hide All Steps

Hint : This problem can be worked in the same manner as the previous two or because each term is a logarithm an easier solution would be to use the fact that, ${\mbox{If }}{\log_b}x = {\log _b}y{\mbox{ then }}x = y$
Start Solution

While we could use the same method we used in the previous couple of examples to solve this equation there is an easier method. Because each of the terms is a logarithm and it’s all equal to zero we can use the fact that,

${\mbox{If }}{\log_b}x = {\log _b}y{\mbox{ then }}x = y$

So, a quick rewrite of the equation gives,

\begin{align*}2\log \left( z \right) & = \log \left( {7z - 1} \right)\\ \log \left( {{z^2}} \right) & = \log \left( {7z - 1} \right)\end{align*}

Note that in order to use the fact above we need both logarithms to have coefficients of one so we also had to make quick use of one of the logarithm properties to make sure we had a coefficient of one.

Show Step 2

Now all we need to do use the fact and solve for $$z$$.

\begin{align*}{z^2} & = 7z - 1\\ {z^2} - 7z + 1 & = 0\end{align*}

In this case we’ll need to use the quadratic formula to finish this out.

$z = \frac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}} = \frac{{7 \pm \sqrt {45} }}{2} = 0.1459,\,\,\,\,6.8541$ Show Step 3

We’re dealing with logarithms so we need to make sure that we won’t have any problems with any of our potential solutions. In other words, we need to make sure that if we plug either of the two potential solutions into the original equation we won’t end up taking the logarithm of a negative number or zero.

In this case it is pretty easy to plug them in and see that neither of the two potential solutions will result in taking logarithms of negative numbers and so both are solutions to the equation.

In summary then, the solutions to the equation are,

$\require{bbox} \bbox[2pt,border:1px solid black]{{z = \frac{{7 \pm \sqrt {45} }}{2} = 0.1459,\,\,\,\,6.8541}}$

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 than 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 the 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!