Question 172872
Well, as they say in the old country..."you can't that there 'ere"
Let's see how it should go:
a) {{{ln(x-4)+ln(2x) = ln(6)}}} Apply the "product rule" for logarithms...{{{log(M)+log(N) = log(M*N)}}}
{{{ln((x-4)(2x)) = ln(6)}}} Simplify the left side.
{{{ln(2x^2-8x) = ln(6)}}} ...and if {{{log(M) = log(N)}}} then {{{M = N}}}, so...
{{{(2x^2-8x) = 6}}} Subtract 6 from both sides.
{{{2x^2-8x-6 = 0}}} Solve by the quadratic formula:
{{{x = (-(-8)+-sqrt((-8)^2-4(2)(-6)))/2(2)}}}
{{{x = (8+-sqrt(64+48))/4}}}
{{{x = (8+4sqrt(7))/4}}} or {{{x = (8-2sqrt(7))/4}}}
{{{x = 2+sqrt(7)}}} or {{{x = 2-sqrt(7)}}}
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b) {{{e^(2x-7) = 9}}} Take the natural log of both sides.
{{{ln(e^(2x-7)) = ln(9)}}} Apply the "power rule" to the left side.
{{{(2x-7)ln(e) = ln(9)}}} Substitute {{{ln(e) = 1}}} 
{{{2x-7 = ln(9)}}} Evaluate the ln(9).
{{{2x-7 = 2.197}}} Add 7 to both sides.
{{{2x = 9.197}}} Divide both sides by 2.
{{{x = 4.598}}}
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c){{{log((9x+5))-log(3) = 2}}} Apply the "quotient rule" to the left side.
{{{log((9x+5)/3) = 2}}} Rewrite in exponential form.
{{{10^2 = (9x+5)/3}}} Multiply both sides by 3.
{{{300 = 9x+5}}} Subtract 5 from both sides.
{{{295 = 9x}}} Divide both sides by 9.
{{{x = 32.77}}}