How do you solve #e^(3r)+4=59#?

1 Answer
Jan 28, 2017

Answer:

#r=ln55/3=1.336#

Explanation:

#e^(3r)+4=59# can be rewritten as #e^(3r)=59-4=55#

As "#ln#" is natural logarithm to the base #e#,

by taking logarithm of both sides to base #e#,

we get #" "3r=ln55# and

#r=ln55/3=4.00733/3=1.33578# or #1.336#

(using scientific calculator)