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Section 1-9 : Exponential And Logarithm Equations

8. Find all the solutions to \(\ln \left( {y - 1} \right) = 1 + \ln \left( {3y + 2} \right)\). If there are no solutions clearly explain why.

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Hint : Don’t forget about the basic logarithm properties and how they can be used to combine multiple logarithms into a single logarithm.
Start Solution

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,

\[\begin{align*}\ln \left( {y - 1} \right) - \ln \left( {3y + 2} \right) & = 1\\ \ln \left( {\frac{{y - 1}}{{3y + 2}}} \right) & = 1\end{align*}\] Show Step 2

Now all we need to do is exponentiate both sides using \(\bf{e}\) (because we’re working with the natural logarithm) and then solve for \(y\).

\[\begin{align*}{{\bf{e}}^{\ln \left( {\frac{{y - 1}}{{3y + 2}}} \right)}} & = {{\bf{e}}^1}\\ \frac{{y - 1}}{{3y + 2}} & = {\bf{e}}\\ y - 1 & = {\bf{e}}\left( {3y + 2} \right) = 3{\bf{e}}y + 2{\bf{e}}\\ \left( {1 - 3{\bf{e}}} \right)y & = 1 + 2{\bf{e}}\\ y & = \frac{{1 + 2{\bf{e}}}}{{1 - 3{\bf{e}}}} = - 0.8996\end{align*}\] 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 in the 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 our potential solution into the first logarithm we’ll be taking the logarithm of a negative number. The same will be true for the second logarithm and so \(y = - 0.8996\) can’t be a solution.

Because this was our only potential solution we know now that there will be no solutions to this equation.