Question 260514
here is the question
{{{log(x+3)=1-logx}}}
step 12 - add log(x) to both side to get
{{{log(x+3) + log(x) = 1}}}
use rules of logs to get
{{{(log(x^2+3x)) = 1}}}
rewrite as exponents as
{{{10^1 = x^2 + 3x}}}
set = 0 and solve for x as
{{{x^2 + 3x -10 = 0}}}
factor
{{{(x+5)(x-2) = 0}}}
solve
x+5 = 0
x = -5
x-2 = 0
x=2
we can't use x = -5, so the only answer is x = 2.