Question 75293
{{{log((x+3))=1-log(x)}}}
{{{log((x+3)) + log(x) = 1}}}
{{{log((x*(x+3))) = 1}}}
{{{log((x^2 +3x)) = 1}}}
Note that {{{log(10) = 1}}}, so
{{{x^2 + 3x = 10}}}
{{{x^2 + 3x - 10 = 0}}}
{{{(x + 5)(x - 2) = 0}}}
x = -5 
x = +2 
check
{{{log((x+3))=1-log(x)}}}
{{{log((-5+3))=1-log((-5))}}}
x = -5 cannot be a solution, because no exponent with a positive base
can give a negative result
Try x = +2
{{{log((2+3))=1-log(2)}}}
{{{log(5)=1-log(2)}}}
{{{log(5) + log(2) = 1}}}
{{{log((5*2)) = 1}}}
{{{log(10) = 1}}}
{{{1 = 1}}}
OK x = +2 is the answer