A Simple Logarithm Question Solve for $x$:  $\log_2 (2x+8)=3$  

Correct Solution:
$2x+8=2^3$
$2x+8=8$
$2x=0$
$x=0$    

Why doesn't this work:
$\log_2 (2x+8)=3$ 

Expand:
$\log_2(2x)+\log_28=3$
$\log_2(2x)+3=3$
$\log_2(2x)=0$
$2x=2^0$
$2x=1$
$x=1/2$  

Thank you
Note that 
$$\log_2(2x+8)\not=\log_2(2x)+\log_2(8).$$

We have $\log_2(2x)+\log_2(8)=\log_2(2x\color{red}{\times} 8)$.
One more way: to make yr life easier use the property of yhe logsrithm: 

$$
\log_a b = \frac{\ln b}{\ln a}
$$

wher3 ln is a natural log. After a bit of algebra exponentiate both sides.