1. ## Exponential equations questions

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

2. Originally Posted by Paymemoney
$\displaystyle 2^x=9$
has no solution but answer says it is does.
Given the arithmetic is correct

$\displaystyle \Rightarrow x = \log_2(9)$

3. Originally Posted by Paymemoney
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
$\displaystyle y^2-20y+64=0$

Not

$\displaystyle 2y^2...$

4. " 2y^2-20y+64=0 " is incorrect.

2^(2x)-20(2^x)=-64
" Let 2^x be y " is the right direction, but-

2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

As for the rest of the steps, I believe you know them already

5. Originally Posted by Paymemoney
$\displaystyle 2^{2x}-20(2^x)=-64$
Yikes! What are you doing? Keep it simple:

2^2(2^x) - 20(2^x) = -64
4(2^x) - 20(2^x) = -64
-16(2^x) = -64
2^x = -64 / -16
2^x = 4
x = 2

6. lols, Wilmer.

2^2(2^x) = 2^(x+2) is not the same as 2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x)

7. Originally Posted by werepurple
" 2y^2-20y+64=0 " is incorrect.

2^(2x)-20(2^x)=-64
" Let 2^x be y " is the right direction, but-

2^(2x) = (2^x)(2^x) = 2^(x+x) = 2^(2x) = (y)(y) = y^2 ----( not 2y^2 )

As for the rest of the steps, I believe you know them already
oh oops, i think i'll tired it again and i should be able to get it.

8. lol, i didn't ever need to complete the square and i answer my own question.
$\displaystyle (y-16)(y-4)$