Suppose \(f(x)=x+3\text{.}\) What is \(f^{-1}(x)\text{?}\) Well, we know from SectionΒ 4.2 how to do this:
\begin{align*}
y \amp= x+3\\
x \amp= y+3\\
x-3 \amp= y
\end{align*}
So, we have \(f^{-1}(x)=x-3\text{.}\) This makes sense, since subtraction is the opposite of addition, it should be the inverse.
Letβs do another. What is the inverse of \(g(x)=3x\text{?}\) Again, we know how to do this:
\begin{align*}
y \amp= 3x\\
x \amp= 3y\\
\frac{x}{3} \amp= y
\end{align*}
So, we have \(g^{-1}(x)=\frac{x}{3}\text{.}\) This makes sense, since division is the opposite of multiplication, it should be the inverse.
One more: what is the inverse of \(h(x)=x^3\text{?}\) You know the drill by now:
\begin{align*}
y \amp= x^3\\
x \amp= y^3\\
\sqrt[3]{x} \amp= y
\end{align*}
So, we have \(h^{-1}(x)=\sqrt[3]{x}\text{.}\) This makes sense, since the cube root is the opposite of a cube, it should be the inverse.
So what do we do with something like \(3^x\text{?}\) This is an exponential function, so we need something that is the opposite of an exponential. This is where logarithms come in.