How do you solve #-9*e^(3-6r)=-3#?

1 Answer
Oct 8, 2016

Answer:

I found: #r=1/6{3-ln[1/3]}=0,6831#

Explanation:

Let us try rearranging it:
#e^(3-6r)=(-3)/(-9)#
I took the #-9# to the right:
we get:
#e^(3-6r)=1/3#
Take the natural log of both sides:
#ln[e^(3-6r)]=ln[1/3]#
we get:
#3-6r=ln[1/3]#
#6r=3-ln[1/3]#
#r=1/6{3-ln[1/3]}#
if you can use a scientific calculator you'd get:
#r=1/6{3-ln[1/3]}=0,6831#