How do you evalute #Log_ 3 (18)#?

1 Answer
Jan 1, 2016

#log_3(18)=2+log_3(2)approx2.6309#

Explanation:

#log_3(18)=log_3(3^2xx2)#

Use the rule: #log_a(bc)=log_a(b)+log_a(c)#

#=>log_3(3^2)+log_3(2)#

Use the rule: #log_a(a^b)=b#

#=>2+log_3(2)#

This is a simplified answer. However, if you want your answer in decimal form, use a calculator. Since many calculators don't have the capability of finding logarithms with a specific base, use the change of base formula.

#log_a(b)=(log_c(b))/(log_c(a))#

Thus, since the #ln# button is on most calculators (or just the #log# button),

#log_3(2)=ln(2)/ln(3)=log(2)/log(3)approx0.6309#

Thus,

#log_3(18)=2+log_3(2)approx2.6309#