Question 21665
Solve for x:
log{{{3}}}(x)+log{{{3}}}(x-4)+16 = 18 Subtract 16 from both sides.
log{{{3}}}(x)+log{{{3}}}(x-4) = 2 Apply the product rule for logarithms:Log{{{b}}}M+log{{{b}}}N = log{{{b}}}(MN)
log{{{3}}}(x)+log{{{3}}}(x-4) = log{{{3}}}(x^2-4x) = 2 Rewrite in exponential form:
log{{{b}}}x = y means {{{b^y = x}}} so:
log{{{3}}}(x^2-4x) = 2 means {{{3^2 = (x^2-4x)}}} Subtract 3^2 = 9 from both sides.
{{{x^2-4x-9 = 0}}} Solve using the quadratic formula: {{{x = (-b+-sqrt(b^2-4ac))/2a}}}
{{{x = (-(-4)+-sqrt((-4)^2-4(1)(-9)))/2(1)}}}
{{{x = (4+-sqrt(52))/2}}} = {{{(4+-sqrt(4*13))/2}}}
{{{x = 2+-sqrt(13)}}}

The roots are:
{{{x = 2+sqrt(13)}}}
{{{x = 2-sqrt(13)}}}