Change of base

Exponential functions and logarithms: Logarithmic functions

Change of base

Many calculators only have a #\log_{10}(x)# button. Fortunately, you can rewrite a logarithm to an expression of any other logarithm using the calculation rules.

\[\begin{array}{rcl}\log_\blue{a}\left(x\right)&=&\dfrac{\log_\green{b}\left(x\right)}{\log_\green{b}\left(\blue{a}\right)}\end{array}\]

Example

\[\log_\blue{4}\left(x\right)=\frac{\log_{\green{2}}\left(x\right)}{\log_\green{2}\left(\blue{4}\right)}\]

\[\log_\frac{1}{\blue{a}}\left(x\right)=-\log_\blue{a}\left(x\right)\]

Example

\[\log_{\frac{1}{\blue{2}}}\left(x\right)=-\log_\blue{2}\left(x\right)\]

With the change of basis identity, we can also relate logarithms of base #\frac{1}{\blue{a}}# with those of base #\blue{a}#.

With the help of these rules we can solve even more logarithmic equations.

Solve the equation for #x#.
\[\log_4\left(x+3\right)=\log_4\left(4\right)-\log_{\frac{1}{4}}\left(x\right)\]

Give your answer in the form #x=\ldots#. Simplify the number at the dots as much as possible.

#x=1#

\(\begin{array}{rcl}
\log_4\left(x+3\right)&=&\log_4\left(4\right)-\log_{\frac{1}{4}}\left(x\right)\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
\log_4\left(x+3\right)&=&\log_4\left(4\right)+\log_{4}\left(x\right)\\
&&\phantom{xxx}\blue{\log_{\frac{1}{a}}\left(b\right)=-\log_a\left(b\right)}\\
\log_4\left(x+3\right)&=&\log_4\left(4 x\right)\\
&&\phantom{xxx}\blue{\log_a\left(b\right)+\log_a\left(c\right)=\log_a\left(b\cdot c\right)}\\
x+3&=&4 x\\
&&\phantom{xxx}\blue{\log_a\left(b\right)=\log_a\left(c\right)\text{ gives }b=c}\\
-3x+3&=&0\\
&&\phantom{xxx}\blue{\text{moved all terms of }x\text{ to the left}}\\
-3x&=&-3\\
&&\phantom{xxx}\blue{\text{moved all constant terms to the right}}\\
x&=&1\\
&&\phantom{xxx}\blue{\text{both sides divided by }-3}
\end{array}\)

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