Question 193653
In general:
{{{log(a) + log(b) = log(ab)}}} so,
{{{log((x+1))+log((x-2))= log((x+1)*(x-2))}}}
also, {{{1 = log(10)}}}
{{{log((x+1)*(x-2)) = log(10)}}}
If the logs are equal, the antilogs are equal
{{{(x+1)*(x-2) = 10}}}
{{{x^2 + x - 2x - 2 - 10 = 0}}}
{{{x^2 - x - 12 = 0}}}
{{{(x - 4)*(x + 3) = 0}}}
{{{x = 4}}}
{{{x = -3}}} this result doesn't work, becase it
gives me a {{{log(-5)}}} and no log can give a 
negative answer
check:
{{{x = 4}}}
{{{log(x+1)+log(x-2)= 1}}}
{{{log(5) + log(2) = 1}}}
{{{.6989700043 + .3010299957 = 1}}}
{{{.9999999997 = 1}}} (inaccuracy is due to rounding off)
OK