Question 129545
{{{log(3^4) - log(3^x) = 2}}}
Therre are a couple of general rules that help
{{{log(a*b) = log(a) + log(b)}}}
and also
{{{log(a/b) = log(a) - log(b)}}}
I can work this 2nd equation backwards
{{{log(a) - log(b) = log(a/b)}}}
{{{log(3^4) - log(3^x) = log((3^4)/(3^x))}}}
{{{log((3^4)/(3^x)) = 2}}}
I know that {{{log(100) = 2}}}, so
{{{log(100) = log((3^4)/(3^x))}}} 
and it follows that
{{{100 = (3^4)/(3^x)}}}
{{{100 = 81 / 3^x}}}
{{{3^x = .81}}}
take the log of both sides
{{{x*log(3) = log(.81)}}}
{{{x = log(.81) / log(3)}}}
{{{x = -.1918}}}
check answer
{{{log(3^4) - log(3^x) = 2}}}
{{{log(81) - log(3^(-.1918)) = 2}}}
{{{1.9085 - log(.81) = 2}}}
{{{1.9085 - (-.0915) = 2}}}
{{{1.9085 + .0915 = 2}}}
{{{2 = 2}}}
OK