### Exponential functions and logarithms: Logarithmic functions

### Isolating variables

Using what we learned when rewriting logarithms, we can rewrite exponential functions to a function of the form #x=\ldots#. We call this isolating #x#.

We can rewrite the function #y=5\cdot 4^{x-1}# as an equation of the form #x=\blue{a}+\log_{\green{b}}\left(\purple{c}\cdot y\right)#.

\[\begin{array}{rcl}5\cdot 4^{x-1}&=&y\\&& \blue{\text{original equation}}\\4^{x-1}&=&\frac{y}{5}\\&&\blue{\text{both sides divided by }5}\\x-1&=&\log_4\left(\frac{y}{5}\right)\\&&\blue{a^b=c\text{ gives }b=\log_a\left(c\right)}\\x&=&\log_4\left(\frac{y}{5}\right)+1\\&&\blue{\text{add }1\text{ on both sides}}\end{array}\]

We can also isolate #x# from more difficult functions, as we can see in the following examples.

Isolate #x# from the following equation.

\[y=6+5^{2.1\cdot x+3}\]

\[y=6+5^{2.1\cdot x+3}\]

#x=# #\frac{\log_{5}\left(y-6\right)-3}{2.1}#

\(\begin{array}{rcl}

y&=&6+5^{2.1\cdot x+3}\\

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

5^{2.1\cdot x+3}&=&y-6\\

&&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\

2.1\cdot x+3&=&\log_{5}\left(y-6\right)\\

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

2.1\cdot x&=&\log_{5}\left(y-6\right)-3\\

&&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\

x&=&\dfrac{\log_{5}\left(y-6\right)-3}{2.1}\\

&&\phantom{xxx}\blue{\text{both sides divided by }2.1}

\end{array}\)

\(\begin{array}{rcl}

y&=&6+5^{2.1\cdot x+3}\\

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

5^{2.1\cdot x+3}&=&y-6\\

&&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\

2.1\cdot x+3&=&\log_{5}\left(y-6\right)\\

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

2.1\cdot x&=&\log_{5}\left(y-6\right)-3\\

&&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\

x&=&\dfrac{\log_{5}\left(y-6\right)-3}{2.1}\\

&&\phantom{xxx}\blue{\text{both sides divided by }2.1}

\end{array}\)

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