Solve for x where loge(4+3x) = 2

Solve for x where loge(4+3x) = 2?

answer is 1/3 (e^2 - 4)

i wonder how to do it

ln (4+3x) = 2

e^2 = (4+3x)

3x = e^2 - 4

x = (e^2 -4) /3

answer

raise both sides to a power of "e":

4+3x = e^2 (e^loge cancels)

3x = e^2 - 4 (subtract 4 from both sides)

x = 1/3 (e^2 - 4) (divide both sides by 3)

To remove log operate e on both sides
then,
4+3x=e^2
3x=e^2-4
x=1/3(e^2-4)

ln(4+3x) = 2
e^2=4+3x
e^2-4=3x
x=(e^2-4)/3
x=(1/3)(e^2-4)