I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*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.

### Solving Logarithm Equations

Solving logarithm equations are similar to exponential equations. First, we isolate the logarithm on one side by itself with a coefficient of one. Then we use Property 4 from the Logarithm Properties section with an appropriate choice of \(b\). In choosing the appropriate \(b\), we need to remember that the \(b\) MUST match the base on the logarithm!

Solve each of the following equations. Show All Solutions Hide All Solutions

- \(4\log \left( {1 - 5x} \right) = 2\)

Show SolutionThe first step is to divide by the 4, then we’ll convert to an exponential equation.

\[\begin{align*}\log \left( {1 - 5x} \right) & = \frac{1}{2}\\ {10^{\log \left( {1 - 5x} \right)}} & = {10^{\frac{1}{2}}}\\ 1 - 5x & = {10^{\frac{1}{2}}}\end{align*}\]Note that since we had a common log in the original equation we were forced to use a base of 10 in the exponential equation. Once we’ve used Property 4 to simplify the equation we’ve got an equation that can be solved.

\[\begin{align*}1 - 5x & = {10^{\frac{1}{2}}}\\ - 5x & = - 1 + {10^{\frac{1}{2}}}\\ x & = - \frac{1}{5}\left( { - 1 + {{10}^{\frac{1}{2}}}} \right)\\ x & = -0.4324555320\end{align*}\]Now, with exponential equations we were done at this point, but we’ve got a little more work to do in this case. Recall the answer to the domain of a logarithm (the answer to Problem 9 in the Logarithm Properties section). We can’t take the logarithm of a negative number or zero.

This does not mean that \(x = -0.4324555320\) can’t be a solution just because it’s negative number! The question we’ve got to ask is this: does this solution produce a negative number (or zero) when we plug it into the logarithms in the original equation? In other words, is \(1 - 5x\) negative or zero if we plug \(x = -0.4324555320\) into it? Clearly, (I hope…) \(1 - 5x\) will be positive when we plug \(x = -0.4324555320\) in.

Therefore, the solution to this is \(x = - 0.4324555320\).

Note that it is possible for logarithm equations to have no solutions, so if that should happen don’t get to excited!

- \(3 + 2\ln \left( {\frac{x}{7} + 3} \right) = - 4\)

Show SolutionThere’s a little more simplification work to do initially this time, but it’s not too bad.

\[\begin{align*}2\ln \left( {\frac{x}{7} + 3} \right) & = - 7\\ \ln \left( {\frac{x}{7} + 3} \right) & = - \frac{7}{2}\\ {{\bf{e}}^{\ln \left( {\frac{x}{7} + 3} \right)}} & = {{\bf{e}}^{ - \frac{7}{2}}}\\ \frac{x}{7} + 3 & = {{\bf{e}}^{ - \frac{7}{2}}}\end{align*}\]Now, solve this.

\[\begin{align*}\frac{x}{7} + 3 & = {{\bf{e}}^{ - \frac{7}{2}}}\\ \frac{x}{7} & = - 3 + {{\bf{e}}^{ - \frac{7}{2}}}\\ x & = 7\left( { - 3 + {{\bf{e}}^{ - \frac{7}{2}}}} \right)\\ x & = {\rm{ - 20}}.78861832\end{align*}\]I’ll leave it to you to check that \(\frac{x}{7} + 3\) will be positive upon plugging \(x = - 20.78861832\) into it and so we’ve got the solution to the equation.

- \(2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) = 2\)

Show SolutionThis one is a little different from the previous two. There are two logarithms in the problem. All we need to do is use Properties 5 – 7 from the Logarithm Properties section to simplify things into a single logarithm then we can proceed as we did in the previous two problems.

The first step is to get coefficients of one in front of both logs.

\[\begin{align*}2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) & = 2\\ \ln \left( x \right) - \ln \left( {1 - x} \right) & = 2\end{align*}\]Now, use Property 6 from the Logarithm Properties section to combine into the following log.

\[\ln \left( {\frac{x}{{1 - x}}} \right) = 2\]Finally, exponentiate both sides and solve.

\[\begin{align*}\frac{x}{{1 - x}} & = {{\bf{e}}^2}\\ x & = {{\bf{e}}^2}\left( {1 - x} \right)\\ x & = {{\bf{e}}^2} - {{\bf{e}}^2}x\\ x\left( {1 + {{\bf{e}}^2}} \right) & = {{\bf{e}}^2}\\ x & = \frac{{{{\bf{e}}^2}}}{{1 + {{\bf{e}}^2}}}\\ x & = 0.8807970780\end{align*}\]Finally, we just need to make sure that the solution, \(x = 0.8807970780\), doesn’t produce negative numbers in both of the original logarithms. It doesn’t, so this is in fact our solution to this problem.

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

Show SolutionThis one is the same as the last one except we’ll use Property 5 to do the simplification instead.

\[\begin{align*}\log x + \log \left( {x - 3} \right) & = 1\\ \log \left( {x\left( {x - 3} \right)} \right) & = 1\\ {10^{\log \left( {{x^2} - 3x} \right)}} & = {10^1}\\ {x^2} - 3x & = 10\\ {x^2} - 3x - 10 & = 0\\ \left( {x - 5} \right)\left( {x + 2} \right) & = 0\end{align*}\]So, potential solutions are \(x = 5\) and \(x = - 2\). Note, however that if we plug \(x = - 2\) into either of the two original logarithms we would get negative numbers so this can’t be a solution. We can however, use \(x = 5\).

Therefore, the solution to this equation is \(x = 5\).

It is important to check your potential solutions in the

**original**equation. If you check them in the second logarithm above (after we’ve combined the two logs) both solutions will appear to work! This is because in combining the two logarithms we’ve actually changed the problem. In fact, it is this change that introduces the extra solution that we couldn’t use!So, be careful in finding solutions to equations containing logarithms. Also, do not get locked into the idea that you will get two potential solutions and only one of these will work. It is possible to have problems where both are solutions and where neither are solutions.