Question 806443
Log base 21 (x+4) = 1- log base 21 x
we can write it:
{{{log(21,(x+4))}}} = 1 - {{{log(21,(x))}}} 
Get the logs on the same side of the equation
{{{log(21,(x+4))}}} + {{{log(21,(x))}}} = 1
adding logs, of the same base, is multiply so we have
{{{log(21,(x(x+4)))}}} = 1
The exponent equiv of logs
x(x+4) = {{{21^1}}}
or just
x(x+4) = 21
a quadratic equation
x^2 + 4x - 21 = 0
Factors to
(x+7)(x-3) = 0
our solution has to be positive
x = 3