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### Section 6.4 : Solving Logarithm Equations

5. Solve the following equation.

${\log _2}\left( {x + 1} \right) - {\log _2}\left( {2 - x} \right) = 3$

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Hint : If we can reduce all the logarithms to a single logarithm it would be quite easy to convert to exponential form. Also, don’t forget that the values with get when we are done solving don’t always correspond to actual solutions so be careful!
Start Solution

First let’s notice that we can combine the two logarithms on the left side to get,

${\log _2}\left( {\frac{{x + 1}}{{2 - x}}} \right) = 3$ Show Step 2

Now, we can easily convert this to exponential form.

$\frac{{x + 1}}{{2 - x}} = {2^3} = 8$ Show Step 3

Now all we need to do is solve the equation from Step 2 and that is an equation that we know how to solve. Here is the solution work.

\begin{align*}\frac{{x + 1}}{{2 - x}} & = 8\\ x + 1 & = 8\left( {2 - x} \right) = 16 - 8x\\ 9x & = 15\hspace{0.25in} \to \hspace{0.25in}x = \frac{{15}}{9} = \frac{5}{3}\end{align*} Show Step 4

As the final step we need to take the number from the above step and plug it into the original equation from the problem statement to make sure we don’t end up taking the logarithm of zero or negative numbers!

Here is the checking work for the number.

$x = \frac{5}{3}:$ \begin{align*}{\log _2}\left( {\frac{5}{3} + 1} \right) - {\log _2}\left( {2 - \frac{5}{3}} \right) & = 3\\ {\log _2}\left( {\frac{8}{3}} \right) - {\log _2}\left( {\frac{1}{3}} \right) & = 3\hspace{0.25in}\,\,\,\,\,{\mbox{OKAY}}\end{align*}

In this case, the number did not produce negative numbers in the logarithms so it is in fact a solution (won’t happen with every problem so don’t always expect this to happen!).

Therefore, the solution to the equation is then : $$\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{5}{3}}}$$.