Question

How do you solve `e^(0.6x-5)+9=18`?

Answer 1

I got `x=12`

Explanation:

We can first take the `9` to the right:
`e^(0.6x-5)=18-9`
`e^(0.6x-5)=9`

we then take the natural log, `ln` of both sides:
`ln[e^(0.6x-5)]=ln(9)`

On the left the `ln` and the exponential cancel each other and we end up with:

`0.6x-5=ln(9)`

rearrange to isolate `x`:

`x=(ln(9)+5)/0.6=11.99~~12`