Hi

i have tried to solve these equations but still cannot get the correct answer:

The following is what i have done:

$\displaystyle 2^{2x}-20(2^x)=-64$

make $\displaystyle y=2^x$

$\displaystyle 2y^2-20y+64=0$

$\displaystyle 2(y^2-10y+32)$

$\displaystyle 2(y^2-10y-(\frac{10}{2})^2+32-25)$

$\displaystyle 2(y-5)^2+7$

$\displaystyle 2(y-5+\sqrt{7})(y-5-\sqrt{7})$

then:

$\displaystyle 2(y-5+\sqrt{7})$

$\displaystyle 2y-10+2\sqrt{7}=0$

$\displaystyle 2y+2\sqrt{7}=10$

$\displaystyle 2y+\sqrt{7}=5$

$\displaystyle 2y=25-7$

$\displaystyle y=\frac{18}{2}$

$\displaystyle 2^x=9$

has no solution but answer says it is does.

$\displaystyle 2(y-5-\sqrt{7})$

$\displaystyle 2y-10-2\sqrt{7}=0$

$\displaystyle 2y-2\sqrt{7}=10$

$\displaystyle 2y=25+7$

$\displaystyle y=\frac{32}{2}$

$\displaystyle 2^x=16$

$\displaystyle 2^x=16$

$\displaystyle x=4$

P.S