Given that $\log_8(x+2)$ $+$ $\log_8y$ $=$ $z-\frac{1}{3}$ and $\log_2(x-2)$ $-$ $\log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $

Question

Given that $\log_8(x+2)$ $+$ $\log_8y$ $=$ $z-\frac{1}{3}$ and $\log_2(x-2)$ $-$ $\log_2y$ = $2z+1$, Show that $x^2 = 32 ^z + 4 $. Any hints is appreciated. Thanks.

Answer 1

Well, $\log_8(x+2) + \log_8 y = \log_8y(x+2)= z-\frac 13$. And $\log 2{x-2} + \log_2 y = \log_2 \frac {x-2}y=2z + 1$

So $y(x+2) = 8^{z-\frac 13} = \frac {8^z}{2}$

And $ \frac {x-2}y = 2^{2z+1}= 2*2^{2z}$.

Hmmph.... Normally I'd solve for $y$ in terms of $x$ for one and substitute that value in the other and solve for $x$ but...

Just multiply the two together to get

$y(x+2)(\frac {x-2}y) =\frac {8^z}2*(2*2^{2z})$

$(x + 2)(x -2) = 8^z*2^{2z}$

$x^2 -4 = 2^{3z}2^{2z} = 2^{5z} = 32^z$.

$x^2 = 32^z + 4$

Answer 2

Hints:

$\log a+\log b=\log(ab)$

$\log_8c=\log_2c/\log_2 8,$ so $3\log_8 c=\log_2 c$

$2^5=32$

Let me know if you need more help. Here's an answer:

Given that $\log_8(x+2)$ $+$ $\log_8y$ $=$ $z-\frac{1}{3}$ and $\log_2(x-2)$ $-$ $\log_2y$ = $2z+1$, we have $\log_8((x+2)y)=z-\frac13$ and $\log_2(\frac{x-2}y)=2z+1$, i.e., $\log_2((x+2)y)=3z-1.$ Thus, $\frac{x-2}y=2^{2z+1}$ and $(x+2)y=2^{3z-1}.$ Multiplying these equations, $x^2-4=2^{5z}$, so $x^2=2^{5z}+4=32^z+4$.