Solve for $x$: $$\log_2 \frac{2x+8}{x+2} +\log_2\frac{x+2}{x-5}=3$$
I see that both logarithms have the same base, 2, so I can use the product rule of logarithms to combine them into one. I get $$\log_2\left(\frac{2x+8}{x+2}\cdot \frac{x+2}{x-5}\right)=3$$ Simplifying the fractions, I see that the $(x+2)$ factors cancel out, leaving $$\log_2\frac{2x+8}{x-5}=3$$ Now I can exponentiate both sides with base 2 to get rid of the logarithm. I get $$\frac{2x+8}{x-5}=2^3$$ Multiplying both sides by $(x-5)$, I get $$2x+8=8(x-5)$$ Expanding the right side, I get $$2x+8=8x-40$$ Subtracting $2x$ from both sides, I get $$8=6x-40$$ Adding $40$ to both sides, I get $$48=6x$$ Dividing both sides by $6$, I get $$x=8$$ 