# Algebra

• October 9th 2008, 06:36 PM
xwrathbringerx
Algebra
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?
• October 10th 2008, 09:25 AM
CaptainBlack
Quote:

Originally Posted by xwrathbringerx
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?

$z^x=y^{2x}$ implies $z=y^2$

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

so

$x+y+z=6$

can be written:

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

so now just check the candidate.

CB