Question 4867
i take it you mean {{{ln (x^3/(x-3)^2)}}}? If so, it becomes


{{{ln(x^3) - ln(x-3)^2}}}
{{{3ln(x) - 2ln(x-3)}}}


solve {{{log10 (x-30)= 3- log10 x}}}..I shall leave out the base, for ease of reading --> {{{log(x-30)= 3 - log(x)}}}.


{{{log(x-30) + log(x) = 3}}}
{{{log(x(x-30)) = 3}}}
{{{log(x^2-30x) = 3}}}
{{{x^2-30x = 10^3}}}
{{{x^2-30x = 1000}}}
{{{x^2-30x-1000 = 0}}}

(x+20)(x-50) = 0


so x+20 = 0 OR x-50=0


so, x=-20 or x=50


in terms of logs, you cannot have negatives, so x=-20 will not be allowed, so x=50 is your one and only answer.


jon