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# :
\[
2^{x+2}=16
\]
Give your answer in the form #x=\ldots# and do not use any exponents in your answer.
#x=2#
\(\begin{array}{rcl}
2^{x+2}&=&16\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
x+2&=&\log_{2}\left(16\right)\\
&&\phantom{xxx}\blue{a^x=b\text{ gives }x=\log_a\left(b\right)}\\
x+2&=&4\\
&&\phantom{xxx}\blue{\text{continue on the logarithm}}\\
x&=&2\\
&&\phantom{xxx}\blue{\text{move constant terms to the right}}\\
\end{array}\)
\(\begin{array}{rcl}
2^{x+2}&=&16\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
x+2&=&\log_{2}\left(16\right)\\
&&\phantom{xxx}\blue{a^x=b\text{ gives }x=\log_a\left(b\right)}\\
x+2&=&4\\
&&\phantom{xxx}\blue{\text{continue on the logarithm}}\\
x&=&2\\
&&\phantom{xxx}\blue{\text{move constant terms to the right}}\\
\end{array}\)
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