Question 137134
Solve for x:
{{{2log((x-2)) = log((x-25))}}} First apply the "power rule" for logarithms to the left side: {{{a*log((b)) = log((b))^a}}}
{{{log((x-2))^2 = log((x-25))}}} Now apply the property: If {{{log((a)) = log((b))}}}, then {{{a = b}}}
{{{(x-2)^2 = x-25}}} Expand the left side.
{{{x^2-4x+4 = x-25}}} Subtract x from both sides.
{{{x^2-5x+4 = -25}}} Add 25 to both sides.
{{{x^2-5x+29 = 0}}} Solve using the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}}
{{{x = (-(-5)+-sqrt((-5)^2-4(1)(29)))/2(1)}}} Simplify.
{{{x = (5+-sqrt(25-116))/2}}}
{{{x = (5+-sqrt(-91))/2}}}
{{{x = (5/2)+(1/2)sqrt(91)i}}} or {{{x = (5/2)-(1/2)sqrt(91)i}}}
OOPS!!! I misread the problem! It should be:
{{{2log((x))-2 = log((x-25))}}} Apply the power rule to the left side:
{{{log((x^2)) - 2 = log((x-25))}}} Add 2 to both sides.
{{{log((x^2)) = log((x-25))+2}}}  Subtract {{{log((x-25))}}} from both sides.
{{{log((x^2))-log((x-25)) = 2}}} Apply the quotient rule to the left side:{{{log((M))-log((N)) = log((M/N))}}}
{{{log(((x^2)/(x-25))) = 2}}} Rewrite this in exponential form: {{{y = Log[b](x) }}}--->{{{x = b^y}}} The base is 10 for commom logarithms:
{{{10^2 = (x^2)/((x-25))}}} Multiply both sides by (x-25)
{{{100(x-25) = x^2}}} Simplify.
{{{100x-2500 = x^2}}} Subtract 100x from both sides and add 2500 to both sides.
{{{x^2-100x+2500 = 0}}} Factor.
{{{x-50)(x-50) = 0}}} Apply the zero products rule.
{{{x-50 = 0}}} and {{{x-50 = 0}}} and so...
{{{x = 50}}}