Question 507520
Solve:
{{{Log(3x+4) = 2Log(x)}}} Apply the "power rule" to the right side.
{{{Log(3x+4) = Log(x^2)}}} Now if {{{Log(M) = Log(N)}}} then {{{M = N}}} so...
{{{3x+4 = x^2}}} Rearrange as:
{{{x^2-3x-4 = 0}}} Solve by factoring:
{{{(x+1)(x-4) = 0}}} Apply the "zero product" rule:
{{{x+1 = 0}}} or {{{x-4 = 0}}} so that...
{{{x = -1}}} or {{{x = 4}}} 
Now you must check the results for "extraneous" or "invalid" values:
Try {{{x = -1}}}
{{{Log(3x+4) = 2Log(x)}}} Substitute {{{x = -1}}}
{{{Log(3(-1)+4) = 2Log(-1)}}}
{{{Log(1) = 2Log(-1)}}}
{{{0 = 2(1.364376)i}}} This is not valid!
Try {{{x = 4}}}
{{{Log(3x+4) = 2Log(x)}}} Substitute {{{x = 4}}}
{{{Log(16) = 2Log(4)}}}
{{{1.20411998 = 2(0.60205999)}}}
{{{1.20411998 = 1.20411998}}} This is valid.
Solution is {{{x = 4}}}