## 9.4 Solving equations with exponents

## Worked example 9.13: Solving equations with exponents

Solve for \(b\):

\[b^{2} = 169\]### First method (use the inverse operation).

The inverse operation when a variable is being squared like \(b^{2}\) is \(\sqrt{b}\) (the square root of \(b\)).

\[\begin{align} b^{2} &= 169 \\ \sqrt{b^{2}} &= \sqrt{169} \\ b &= \pm 13 \end{align}\]The \(\pm\) before the answer means that the answer could be negative \(13\) or positive \(13\) because \(( - 13)^{2}\) and \((13)^{2}\) both give us \(169\).

### Second method (use laws of exponents).

We could also use a different method. We know that if \(a^{x} = b^{x}\), then \(a = b\). We can use this property to solve the given equation.

\[\begin{align} b^{2} &= 169 \\ b^{2} &= {( \pm 13)}^{2} \\ \therefore b &= \pm 13 \end{align}\]## Worked example 9.14: Using laws of exponents

Solve for \(b\):

\[(b - 7)^{2} = 169\]### Use the inverse operation.

This can be tricky if you do not recognise that the left- and right-hand sides of the equation can be simplified by taking the square root on both sides. Remember that what you do to one side, you must do to the other side, in order to keep the equation balanced. Here we will have two answers because we are using the square root.

\[\begin{align} (b - 7)^{2} &= 169 \\ \sqrt{(b - 7)^{2}} &= \pm \sqrt{169} \end{align}\]Your answer can be a positive or a negative. Both values squared will give you \(169\).

### Use the positive value of the root to find the first solution.

\[\begin{align} b - 7 &= + \left( \sqrt{169} \right) \\ b - 7 &= 13 \\ b &= 13 + 7 \\ b &= 20 \end{align}\]### Use the negative value of the square root to find the second solution.

\[\begin{align} b - 7 &= - \left( \sqrt{169} \right) \\ b - 7 &= - 13 \\ b &= - 13 + 7 \\ b &= - 6 \end{align}\]So, the two solutions to the equation are \(b = - 6\) and \(b = 20\).