Question 60837
You can always rewrite the logarithmic equation as an equivalent exponential equation:

{{{logb(M) = N}}} ---> {{{b^N = M}}} where b is the base of the logarithm.

a) {{{log3(x) = 8}}} ---> {{{3^8 = x}}} ---> {{{6561 = x}}}

b) {{{27^((2x+2)) = 81^((x-3))}}} but {{{27 = 3^3}}} and {{{81 = 3^4}}}, so:
{{{(3^3)^((2x+2)) = (3^4)^((x-3))}}} Multiplying the exponents on both sides.
{{{3^((6x+6)) = 3^((4x-12))}}} but, if {{{x^n = x^m}}} then {{{n = m}}}...the exponents are equal.
{{{6x+6 = 4x-12}}} Simplify and solve for x.
{{{2x = -18}}}
{{{x = -9}}}

c){{{log2(3x+2) = 3}}}  Similar to problem a) Rewrite in exponential form.
{{{2^3 = 3x+2}}}
{{{8 = 3x+2}}} Subtract 2 from both sides.
{{{3x = 6}}} Divide both sides by 3.
{{{x = 2}}}

d) {{{e^(3x) = 80}}} Take the natural log (because of the e) of both sides.
{{{3xln(e) = ln(80)}}} Because {{{ln(e)^x = xln(e)}}}.  Recall that {{{ln(e) = 1}}}
{{{3x = 4.3820}}}
{{{x = 1.4607}}}