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We’ve got two logarithms on one side so we’ll combine those, drop the logarithms and then solve.

\[\begin\ln \left( \right) & = \ln x\\ \frac & = x\\ 10 & = x\left( \right)\\ 10 & = 7x - \\ - 7x 10 & = 0\\ \left( \right)\left( \right) & = 0\hspace \Rightarrow \hspacex = 2,\,\,x = 5\end\] We’ve got two possible solutions to check here.

This will be important down the road and so we can’t forget that.

Now, let’s start off by looking at equations in which each term is a logarithm and all the bases on the logarithms are the same.

We are excluding it because once we plug it into the original equation we end up with logarithms of negative numbers.

It is possible to have negative values of \(x\) be solutions to these problems, so don’t mistake the reason for excluding this value.Also, along those lines we didn’t take \(x = 6\) as a solution because it was positive, but because it didn’t produce any negative numbers or zero in the logarithms upon substitution.It is possible for positive numbers to not be solutions.\(x = 2 :\) \[\begin\ln 10 - \ln \left( \right) & = \ln 2\ \ln 10 - \ln 5 & = \ln 2\end\] This one is okay.\(x = 5 :\) \[\begin\ln 10 - \ln \left( \right) & = \ln 5\ \ln 10 - \ln 2 & = \ln 5\end\] This one is also okay.So, we saw how to do this kind of work in a set of examples in the previous section so we just need to do the same thing here.It doesn’t really matter how we do this, but since one side already has one logarithm on it we might as well combine the logs on the other side.Now we need to take a look at the second kind of logarithmic equation that we’ll be solving here.This equation will have all the terms but one be a logarithm and the one term that doesn’t have a logarithm will be a constant.In this case we will use the fact that, \[x = yx = y\] In other words, if we’ve got two logs in the problem, one on either side of an equal sign and both with a coefficient of one, then we can just drop the logarithms. With this equation there are only two logarithms in the equation so it’s easy to get on one either side of the equal sign.We will also need to deal with the coefficient in front of the first term.

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