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### 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.

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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 2Now 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 3We’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!