zmudz one year ago Let $$x, y,$$ and $$z$$ be positive real numbers that satisfy $$2 \log_x (2y) = 2 \log_{2x} (4z) = \log_{2x^4} (8yz) \neq 0.$$ The value of $$xy^5 z$$ can be expressed in the form $$\frac{1}{2^{p/q}}$$, where $$p$$ and $$q$$ are relatively prime positive integers. Find $$p + q$$.

1. Loser66

Just curious :) Which class is it?

2. zmudz

precalc

3. Loser66

serious??? no way!!

4. freckles

well what I did was set each one of those things to u and then I solved for u and then went back and solved for y then z finally went back and found x. then plugged into xy^5z and did get something in the form of 2^(-p/q) then I just did p+q so first step: $u=\log_x(4y^2) \\ x^u=4y^2$ $u=\log_{2x}(16z^2) \\ x^{u}=2^{4-u}z^2 \\$ $u=\log_{2x^4}(8yz)=u \\ x^{u}=2^{\frac{3-u}{4}}(yz)^\frac{1}{4}$ so I setup a system of equations $2^2y^2=2^{4-u}z^2 \\ 2^{4-u}z^2=2^{3-u}{4}(yz)^\frac{1}{4} \\ 2^2y^2=2^{\frac{2-u}{4}}(yz)^\frac{1}{4} \\$ Eventually got: $z^2=2^{-2+u}y^2 \text{ then also } \\ y^2=2^\frac{-12-u}{6} \text{ and then found } z=2^{\frac{5u-24}{6}}$ $2^{4-u}2^{5u-24}{6}=2^\frac{3-u}{4}2^\frac{-12-u}{48}2^\frac{5u-24}{48} \\ \text{ use law of exponents to get the equation } \\ 2^{4-u+\frac{5u-24}{6}}=2^{\frac{3-u}{4}+\frac{-12-u}{48}+\frac{5u-24}{48}} \\ \text{ set the exponents on both sides equal \to each other } \\ \text{ you can find } u \text{ now }$ I plug this u into $z^2=2^{-2+u}y^2 \text{ then also } \\ y^2=2^\frac{-12-u}{6} \text{ and then found } z=2^{\frac{5u-24}{6}}$ to find y and z. Once I obtained u,y,z... I found x by going back to one of the equations I started with like $x^{u}=4y^2$

5. freckles

Don't know if there is a shorter way.