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.
\[y=50+2^{0.8\cdot x+1}\]
\[y=50+2^{0.8\cdot x+1}\]
#x=# #\frac{\log_{2}\left(y-50\right)-1}{0.8}#
\(\begin{array}{rcl}
y&=&50+2^{0.8\cdot x+1}\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
2^{0.8\cdot x+1}&=&y-50\\
&&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\
0.8\cdot x+1&=&\log_{2}\left(y-50\right)\\
&&\phantom{xxx}\blue{a^b=c \text{ gives } b=\log_a\left(c\right)}\\
0.8\cdot x&=&\log_{2}\left(y-50\right)-1\\
&&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\
x&=&\dfrac{\log_{2}\left(y-50\right)-1}{0.8}\\
&&\phantom{xxx}\blue{\text{both sides divided by }0.8}
\end{array}\)
\(\begin{array}{rcl}
y&=&50+2^{0.8\cdot x+1}\\
&&\phantom{xxx}\blue{\text{the original equation}}\\
2^{0.8\cdot x+1}&=&y-50\\
&&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\
0.8\cdot x+1&=&\log_{2}\left(y-50\right)\\
&&\phantom{xxx}\blue{a^b=c \text{ gives } b=\log_a\left(c\right)}\\
0.8\cdot x&=&\log_{2}\left(y-50\right)-1\\
&&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\
x&=&\dfrac{\log_{2}\left(y-50\right)-1}{0.8}\\
&&\phantom{xxx}\blue{\text{both sides divided by }0.8}
\end{array}\)
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