How do you solve #3*18^(n+8)=1#?

1 Answer
Aug 24, 2016

Answer:

#n = -(8+1/(log_3 2 + 2))#

Explanation:

#3*18^(n+8)=1#. Applying the #log# transformation to both sides

#log_e3+(n+8)log_e 18 = 0# or
#n = -(8log_e 18+log_e 3)/(log_e 18) = -8-log_e3/(log_e 18)# then

#n = -8-log_e3/(log_e2+2log_e 3) = -8-1/(log_e2/log_e3+2#

Finally

#n = -(8+1/(log_3 2 + 2))#