math 30-1: pre-calculus. test

Question

Given: $\log_3 (x-y) = 2\ $ and $\ \log_3 (x+2y) = 4$ find $x + y$.

I have already tried $3^4 = x+2y$ and $3^2 = x-y$.

I am stuck after that.

I am not sure how to approach this question or do it.

Answer 1

You have two equations in your two unknowns X and Y. Find X and find Y and add them.

Answer 2

we have $$3^2=x-y$$ and $$3^4=x+2y$$ by subtracting we get $$3^4-3^2=3y$$ or $$3^3-3=y$$ multiplying the first equation by $2$ and adding both we get $$3x=3^2\cdot 2+3^4$$ thus $$x=6+3^2$$ therefore $$x+y=6+3^3+3^3-3=57$$

Answer 3

$(3^2+3^4) + 3^4 = (2x + y) + (x + 2y) = 3(x+y)$

$x+y = 3+2 \times 3^3 = 57$