Calculus I - Exponential and Logarithm Equations

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Section 1.9 : Exponential And Logarithm Equations

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9. Find all the solutions to \(\log \left( w \right) + \log \left( {w - 21} \right) = 2\). If there are no solutions clearly explain why.

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We need to reduce this down to an equation with a single logarithm and to do that we first should rewrite it a little. Upon doing that we can use the basic logarithm properties to combine the two logarithms into a single logarithm as follows,

\[\begin{align*}\log \left( {w\left( {w - 21} \right)} \right) & = 2\\ \log \left( {{w^2} - 21w} \right) & = 2\end{align*}\] Show Step 2

Now all we need to do is exponentiate both sides using 10 (because we’re working with the common logarithm) and then solve for \(y\).

\[\begin{align*}\log \left( {{w^2} - 21w} \right) & = 2\\ {10^{\log \left( {{w^2} - 21w} \right)}} & = {10^2}\\ {w^2} - 21w & = 100\\ {w^2} - 21w - 100 & = 0\\ \left( {w - 25} \right)\left( {w + 4} \right) & = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}w = - 4,\,\,\,\,w = 25\end{align*}\] 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.

Upon inspection we can quickly see that if we plug in \(w = - 4\) we will be taking a logarithm of a negative number (in both of the logarithms in this case) and so \(w = - 4\) can’t be a solution. On the other hand, if we plug in \(w = 25\) we won’t be taking logarithms of negative numbers and so \(w = 25\) is a solution.

In summary then, the only solution to the equation is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{w = 25}}\).