Solve $\log_2 \frac {18+x}x=2$ for $x$.

Question

Solve the following for $x$:

$$\log_2 \frac {18+x}x=2$$

I am told that the equation implies $\frac {18 + x} x = 4$ (this is the step i didn't understand , Where did $\log_2$ go ?)

Given that, we get the solution. $ x = 6 $ (This step is understood)

Answer 1

That’s pretty much the definition of a logarithm. For any exponential equation $a^b = c$, the logarithmic equation is in form $\log_a c = b$.
$a^b = c \implies\log_a c = b$

Answer 2

$$\log_2 \left(\frac {18+x}x\right)=2$$ $$2^{\log_2 \left(\frac {18+x}x\right)}=2^2$$

use the property $$a^{\log_b c}=c^{\log_b a}$$

$$\left(\frac {18+x}x\right)^{\log_2 2}=4$$

$${\log_2 2}=1$$

$$\frac{18+x}{x}=4$$