How do you solve #e^(-3n)=83#?

1 Answer
May 9, 2017

Answer:

Take the natural log of both sides and solve.

Explanation:

Note that #ln(x)# is the natural logarithm or #log_e(x)#.

First, we can take the natural log of both sides to get #n# out of the exponent:
#ln(e^(-3n))=ln(83)#

Since the #ln# and the #e# cancel, we get:
#-3n=ln(83)#

Dividing #-3# from both sides, we get our value of #n# as:
#n=-ln(83)/3~~-1.473# rounded to three decimal places