A Simple Logarithm Question

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

Answer 1

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

We have $\log_2(2x)+\log_2(8)=\log_2(2x\color{red}{\times} 8)$.

Answer 2

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.