### Exponential functions and logarithms: Logarithmic functions

### Exponential equations

We just practiced solving equations with a form similar to #\blue{a}+\log_{\green{b}} \left(x \right)=\purple{c}#. We'll now have a look at equations of the form #\blue{a}^x=\green{b}#.

\[\blue{a}^x=\green{b}\]

gives

\[x=\log_{\blue{a}}\left(\green{b}\right)\]

**Example**

\[\begin{array}{rcl}\blue{2}^x&=&\green{3}\\x&=&\log_{\blue{2}}\left(\green{3}\right)\end{array}\]

We showed a very simple equation in the above example. However, equations can also be more difficult, as you can see in the examples below.

Solve the equation for #x# :

\[

5^{x+2}=125

\]

Give your answer in the form #x=\ldots# and do not use any exponents in your answer.

#x=1#

\(\begin{array}{rcl}

5^{x+2}&=&125\\

&&\phantom{xxx}\blue{\text{the original equation}}\\

x+2&=&\log_{5}\left(125\right)\\

&&\phantom{xxx}\blue{a^x=b\text{ gives }x=\log_a\left(b\right)}\\

x+2&=&3\\

&&\phantom{xxx}\blue{\text{continue on the logarithm}}\\

x&=&1\\

&&\phantom{xxx}\blue{\text{move constant terms to the right}}\\

\end{array}\)

\(\begin{array}{rcl}

5^{x+2}&=&125\\

&&\phantom{xxx}\blue{\text{the original equation}}\\

x+2&=&\log_{5}\left(125\right)\\

&&\phantom{xxx}\blue{a^x=b\text{ gives }x=\log_a\left(b\right)}\\

x+2&=&3\\

&&\phantom{xxx}\blue{\text{continue on the logarithm}}\\

x&=&1\\

&&\phantom{xxx}\blue{\text{move constant terms to the right}}\\

\end{array}\)

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