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

1 Answer
Sep 16, 2016

Answer:

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#