# Exponential equations questions

• Dec 6th 2009, 07:43 PM
Paymemoney
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
• Dec 6th 2009, 07:49 PM
pickslides
Quote:

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

Given the arithmetic is correct

$\displaystyle \Rightarrow x = \log_2(9)$
• Dec 6th 2009, 07:50 PM
VonNemo19
Quote:

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

$\displaystyle 2y^2...$
• Dec 6th 2009, 07:56 PM
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 :D
• Dec 6th 2009, 07:56 PM
Wilmer
Quote:

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
• Dec 6th 2009, 08:00 PM
werepurple
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)
• Dec 6th 2009, 08:01 PM
Paymemoney
Quote:

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 :D

oh oops, i think i'll tired it again and i should be able to get it.
• Dec 6th 2009, 08:06 PM
Paymemoney
lol, i didn't ever need to complete the square and i answer my own question.
$\displaystyle (y-16)(y-4)$
which will therefore get me x=4 and x=2.

oh well, thanks for spotting the mistake
• Dec 6th 2009, 08:08 PM
Wilmer
Quote:

Originally Posted by werepurple
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)

right...BUT tell me why I got the correct answer (Smirk)
• Dec 6th 2009, 08:10 PM
werepurple
Sheer coincidence, Wilmer. Haha, and that is why there are intersections of lines, curves, and etc; values coincide. ^^