Question 72883
Use these identities to solve

 {{{ylog(x)=log(x)^y}}}
 {{{log(x)+log(y)=log(x*y)}}}

If we let x=2 and y=3 for the 1st identity we get

{{{2*log_[5](3)=log_[5](3^2)=log_[5](9)}}}This gets us halfway to 18.
{{{2*log_[5](3)=2*(0.6826)=1.3652}}}Now plug in the approximate values
{{{log_[5](2)+log_[5](9)=log_[5](2*9)=log_[5](18)}}}Notice how this turns into log(18) using the identities.
{{{0.4307+1.3652=1.7959}}}Plug in remaining values.
Check:
{{{5^1.7959=18.00032}}}It's close enough to work
So {{{log_[5](18)=1.7959}}}Approximately of course