How do you solve #3e ^ { 2x + 1} = 27#?

2 Answers
Gió
Mar 15, 2017

I got: #x=(ln(9)-1)/2=0.598612#

Explanation:

We need to find #x# in: #3e^(2x+1)=27#:
We can rearrange it to write:
#e^(2x+1)=27/3#
#e^(2x+1)=9#
apply the natural log to both sides:
#ln(e^(2x+1))=ln(9)#
and:
#cancel(ln)(cancel(e)^(2x+1))=ln(9)#
#2x+1=ln(9)#
#2x=ln(9)-1#
#x=(ln(9)-1)/2=0.598612#

Anjali G
Mar 15, 2017

#x=ln3-1/2#

Explanation:

#3e^(2x+1)=27#

#e^(2x+1)=9#

#2x+1=ln9#

#2x=ln9-1#

#x=1/2ln9-1/2#

#x=ln(9^(1/2))-1/2#

#x=ln3-1/2#

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