Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$

Question

Question:

Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$

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My attempt: With the first equation

$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$

$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$

$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$

$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$

$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$

Now with the second equation

$$\log_3(x+y)+\log_3(x-y)=1$$

$$\log_3((x+y)(x-y)) = 1 $$

$$ \log_3(x^2-y^2) = 1 $$

$$ x^2-y^2 = 3$$

Now I have 2 equations:

$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$

$$ x^2-y^2 = 3$$

Now I am stuck..

Answer 1

Sorry- I am on my phone in a cafe and this napkin was the best I could do.

Answer 2

The set $$ x^2 - y^2 = 3 $$ in the plane is a hyperbola. You can draw it. What kind of set is $$ x^2 -\frac{5}{2}xy + y^2 = 0, $$ or $$ 2x^2 -5xy + 2y^2 = 0? $$ The second version factors...............

Answer 3

You have the two equations $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\tag 1$$ $$x^2-y^2 = 3\tag 2$$ From $(2)$, $y=\pm \sqrt{x^2-3}$. So, plugging in $(1)$ $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\implies 2(x^2+y^2)=5 x y \implies 2(2x^2-3)=\pm 5x\sqrt{x^2-3}\tag 3$$ Square both sides $$4(2x^2-3)^2=25x^2(x^2-3)$$ Expand and group terms to get $$-9 x^4+27 x^2+36=0$$ which is a quadratic in $t$ if $t=x^2$; its roots are $-1$ and $4$.

I am sure that you can take it from here.

Answer 4

$$\frac{x^2+y^2}{xy}=\frac 52$$ $$y=xt$$ $$\frac {x^2+x^2t^2}{x^2t}=\frac 52$$ $$\frac{1+t^2}{t}=\frac 52$$ $$t=2; t=\frac 12$$ $y=2x \Rightarrow x^2-4x^2=3 x^2-4x^2=3 \Rightarrow x^2=-1$

$y=\frac 12x \Rightarrow x=2y \Rightarrow 4y^2-y^2=3 \Rightarrow y=\pm 1; x =\pm 2$