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.

$y=13+8^{2.5\cdot x+8}$

$x=$ $\frac{\log_{8}\left(y-13\right)-8}{2.5}$

$\begin{array}{rcl} y&=&13+8^{2.5\cdot x+8}\\ &&\phantom{xxx}\blue{\text{the original equation}}\\ 8^{2.5\cdot x+8}&=&y-13\\ &&\phantom{xxx}\blue{\text{swapped and moved the constant term to the right}}\\ 2.5\cdot x+8&=&\log_{8}\left(y-13\right)\\ &&\phantom{xxx}\blue{a^b=c \text{ gives } b=\log_a\left(c\right)}\\ 2.5\cdot x&=&\log_{8}\left(y-13\right)-8\\ &&\phantom{xxx}\blue{\text{moved the constant term to the right}}\\ x&=&\dfrac{\log_{8}\left(y-13\right)-8}{2.5}\\ &&\phantom{xxx}\blue{\text{both sides divided by }2.5} \end{array}$

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