Isolating variables

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)#.

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

Inverse

Isolating the variable #x# from a function is actually the same as determining the inverse of a function #f(x)#. Therefore, we will use the phrase 'give the inverse function #f^{-1}(x)#' in the exercises, as this has the same meaning as 'isolate #x# from the function #f(x)#'. In the above example we use the #y#-notation instead of the notation #f(x)#. The inverse function would then be \[f^{-1}(x)=\log_4\left(\dfrac{x}{5}\right)+1\]

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

Isolate #x# from the following equation.

\[y=24+6^{1.6\cdot x+2}\]

#x=# #\frac{\log_{6}\left(y-24\right)-2}{1.6}#

\(\begin{array}{rcl}
y&=&24+6^{1.6\cdot x+2}\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
6^{1.6\cdot x+2}&=&y-24\\
&&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\
1.6\cdot x+2&=&\log_{6}\left(y-24\right)\\
&&\phantom{xxx}\blue{a^b=c \text{ gives } b=\log_a\left(c\right)}\\
1.6\cdot x&=&\log_{6}\left(y-24\right)-2\\
&&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\
x&=&\dfrac{\log_{6}\left(y-24\right)-2}{1.6}\\
&&\phantom{xxx}\blue{\text{both sides divided by }1.6}
\end{array}\)

New example