# Solving exponential equations

We can solve exponential equation using logs:

Example 1: Find the exact solution of $4e^{2x}=20$.

We begin by dividing both sides by 4 to obtain $e^{2x}=5$. We can then take the natural log of both sides (or use one of the log rules to read this equation) and get $2x=\ln(5)$ .

Finally, $x=\frac{\ln(5)}{2}$.

Example 2: Solve the equation $e^{x}+e^{-x}=2$.

Although it may not appear so just yet, this equation is a hidden quadratic – it is quadratic in e to the x:

$e^{x}+e^{-x}=2\Longrightarrow e^x+\frac{1}{e^x}=2\Longrightarrow \left(e^x\right)^2+1=2e^x$

We obtain the last equation by multiplying both sides of the previous one by e to the x. Setting the right hand side to zero and factorising we get

$\left(e^x\right)^2-2e^x+1=0\Longrightarrow \left(e^x-1\right)^2=0$

We have a repeated root when $e^x=1$,i.e. when x=0.

Example 3 – Solve $2^xe^{3x}=e^5$.

This kind of question appears difficult to students because the combination of e and x makes it look messy. Solving this equation means finding x and currently x appears in the powers. To deal with this we can log both sides of the equation but it is important to remember that both sides must be logged in their entirety:

$\ln\left(2^xe^{3x}\right)=\ln\left(e^5\right)$

We do this because the log rules allow us to change x from being a power inside the log to being a multiplier outside the log. We must first use the log rule that allows us to split the product inside the log to a summation outside the log. Also note that e and ln are inverses of each other and so the equation becomes:

$\ln\left(2^x\right)+\ln\left(e^{3x}\right)=5$

It follows that

$x\ln\left(2\right)+3x=5$

To get x on its own, the left hand side can be factorised to give

$x\left(\ln(2)+3\right)=5$

giving the solution

$x=\frac{5}{\ln(2)+3}$