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?
$\displaystyle z^x=y^{2x}$ 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