Question 79561
Solve for x:
{{{Log[10](x-9)+Log[10](100x) = 3}}} Apply the product rule for logarithms:{{{Log[b](A)+Log[b](B) = Log[b](A*B)}}}
{{{Log[10]((x-9)*(100x)) = 3}}} Simplifying this, we get:
{{{Log[10](100x^2-900x) = 3}}} Now write this in exponential form: {{{Log[b](x) = y}}} means{{{b^y = x}}} Applying this to your problem:
{{{10^3 = 100x^2-900x}}}
{{{1000 = 100x^2-900x}}} Subtract 1000 from both sides.
{{{100x^2-900x-1000 = 0}}} Divide both sides by 100 to simplify this a bit.
{{{x^2-9x-10 = 0}}} Factor this quadratic equation.
{{{(x-10)(x+1) = 0}}} Apply the principle of zero products:
{{{x-10 = 0}}} or {{{x+1 = 0}}}, so...
{{{x = 10}}} or {{{x = -1}}} Discard the negative solution as the log of a negative quantity results in a complex answer.
Your solution is:
{{{x = 10}}}
Check:
{{{Log(10-9)+Log(10*100) = Log(1)+Log(1000)}}}={{{0+3 = 3}}}