Question 80522
Solve for x:
{{{Log[5](3x)+Log[5](x-3) = 1}}} Apply the product rule for logarithms.
{{{Log[5]((3x)(x-3)) = 1}}} Simplify the left side.
{{{Log[5](3x^2-9x) = 1}}} Rewrite in exponential form.
{{{5^1 = 3x^2-9x}}} Subtract 5 from both sides.
{{{3x^2-9x-5 = 0}}} Your work up to this point is commendable!...and you are correct, this quadratic equation is not factorable, so you can use the quadratic formula: {{{x = (-b+-sqrt(b^2-4ac))/2a}}}
{{{x = (-(-9)+-sqrt((-9)^2-4(3)(-5)))/2(3)}}}
{{{x = (9+-sqrt(81+60))/6}}}
{{{x = (9+-sqrt(141))/6}}}
The exact answers are:
{{{x = 3/2+(sqrt(141))/6}}}
{{{x = 3/2-(sqrt(141))/6}}}
The approximate answers are:
{{{x = 3.48}}} To the nearest hundredth.
{{{x = -0.48}}} To the nearest hundredth.