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

Question

1970 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-16T17:36:01

I got $x = 12$ $x=12$

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

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

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

$0.6 x - 5 = ln (9)$ $0.6x-5=ln(9)$

rearrange to isolate $x$ $x$ :

$x = \frac{ln (9) + 5}{0.6} = 11.99 \approx 12$ $x=(ln(9)+5)/0.6=11.99~~12$

How do you solve e^(0.6x-5)+9=18e0.6x−5+9=18e^(0.6x-5)+9=18?