Question

If $\log x+\log (x+3)=1$, find the value of x, where x > 0.

Answer 1

Since, $\log x+\log (x+3)=1$
$\log (x.(x+3\left)\right)=log10$
$\Rightarrow x(x+3)=10$
$\Rightarrow x^2+3x-10=0$
$\Rightarrow x^2+5x-2x-10=0$
$\Rightarrow x(x+5)-2(x+5)=0$
$\Rightarrow (x+5)(x-2)=0$
$\therefore x=2,-5$
but $x>0$
hence, x = 2.