*\[ = \] We now have the same base and a single exponent on each base so we can use the property and set the exponents equal.Doing this gives, \[\beginz & = 2\left( \right)\ z & = 2z 10\ z & = -10\end\] So, after all that work we get a solution of \(z= -10\) .*

*\[ = \] We now have the same base and a single exponent on each base so we can use the property and set the exponents equal.Doing this gives, \[\beginz & = 2\left( \right)\ z & = 2z 10\ z & = -10\end\] So, after all that work we get a solution of \(z= -10\) .*

Because of that all our knowledge about solving equations won’t do us any good.

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.

\[ = ra\] Note that to avoid confusion with \(x\)’s we replaced the \(x\) in this property with an \(a\).

In this part we’ve got some issues with both sides.

First the right side is a fraction and the left side isn’t.

The first thing to do in this problem is to get the same base on both sides and to so that we’ll have to note that we can write both 4 and 8 as a power of 2. \[\begin & = \frac\ & = \frac\end\] It’s now time to take care of the fraction on the right side.

To do this we simply need to remember the following exponent property.

The important part of this property is that we can take an exponent and move it into the front of the term.

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.

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