Solution assignment 09 Logarithmic equations

Solve:

$\log_2(x)+\log_2(x-2)=3$

The domain is:

$x>0$ and $x>2$

thus:

$x>2$

We write the equation as follows:

$\log_2(x)+\log_2(x-2)=\log_2(8)$

$\log_2[(x(x-2)]=\log_2(8)$

$x(x-2)=8$

$x^2-2x-8=0$

$(x-4)(x+2)=0$

The solutions of this equation are:

$x=-2$ or $x=4$

The second solution lies in the domain and is valid. The first solution lies outside the domain and does not satisfy the equation.