If z^x=y^2x, 2^z=2(4^x) and x+y+z=6, then a possible value for y is:

(a) -3/11

(b)11/3

(c)3/11

(d)none of these

Could someone please show me how to find the answer?

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- Oct 9th 2008, 06:36 PMxwrathbringerxAlgebra
If z^x=y^2x, 2^z=2(4^x) and x+y+z=6, then a possible value for y is:

(a) -3/11

(b)11/3

(c)3/11

(d)none of these

Could someone please show me how to find the answer? - Oct 10th 2008, 09:25 AMCaptainBlack
$\displaystyle z^x=y^{2x}$ implies $\displaystyle z=y^2$

$\displaystyle 2^z=2 \times 4^x=2^{2x+1}$ implies that $\displaystyle z=2x+1$ or $\displaystyle x=(z-1)/2=(y^2-1)/2$

so

$\displaystyle x+y+z=6$

can be written:

$\displaystyle (y^2-1)/2+y+y^2=6$

so now just check the candidate.

CB