*So, if we had, \[\] we could use this property as follows. Of course, we are now stuck with a logarithm in the problem and not only that but we haven’t specified the base of the logarithm.The reality is that we can use any logarithm to do this so we should pick one that we can deal with.*

Okay, so we say above that if we had a logarithm in front the left side we could get the \(x\) out of the exponent. We’ll just put a logarithm in front of the left side.

However, if we put a logarithm there we also must put a logarithm in front of the right side.

There are two methods for solving exponential equations.

One method is fairly simple but requires a very special form of the exponential equation.

Again, there really isn’t much to do here other than set the exponents equal since the base is the same in both exponentials.

\[\begin & = 6 - t\ t - 6 & = 0\ \left( \right)\left( \right) & = 0 \hspace \Rightarrow \hspacet = - 3,\,\,t = 2\end\] In this case we get two solutions to the equation.Now that we’ve seen the definitions of exponential and logarithm functions we need to start thinking about how to solve equations involving them.In this section we will look at solving exponential equations and we will look at solving logarithm equations in the next section.This is commonly referred to as taking the logarithm of both sides.We can use any logarithm that we’d like to so let’s try the natural logarithm. Admittedly, it would take a calculator to determine just what those numbers are, but they are numbers and so we can do the same thing here.\[\begin2\left( \right) & = - 3\left( \right)\ 10 - 18x & = - 3x 6\ 4 & = 15x\ x & = \frac\end\] And there is the answer to this part.Now, the equations in the previous set of examples all relied upon the fact that we were able to get the same base on both exponentials, but that just isn’t always possible. \[ = 9\] This is a fairly simple equation however the method we used in the previous examples just won’t work because we don’t know how to write 9 as a power of 7.We need a way to get the \(x\) out of the exponent and luckily for us we have a way to do that.Recall the following logarithm property from the last section.First the right side is a fraction and the left side isn’t.That is not the problem that it might appear to be however, so for a second let’s ignore that.

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