# Solve exponential equations?

• Mar 3rd 2014, 04:11 PM
needhelp96
Solve exponential equations?
Hey I was wondering if you could help me find the answer to some practice questions we were given. I have a quiz tomorrow morning but I'm not even sure if I'm doing it right. Whats the point of practice questions if I don't even have the answers (Crying)

I'm asked to solve exponential equations here are two of the more difficult ones I'm not sure how to do:

Attachment 30287

Could really use a small explanation, or just the steps alone so I can interpret it myself
• Mar 3rd 2014, 04:15 PM
needhelp96
Re: ....
• Mar 3rd 2014, 04:44 PM
romsek
Re: ....
Quote:

Originally Posted by needhelp96
Hey I was wondering if you could help me find the answer to some practice questions we were given. I have a quiz tomorrow morning but I'm not even sure if I'm doing it right. Whats the point of practice questions if I don't even have the answers (Crying)

I'm asked to solve exponential equations here are two of the more difficult ones I'm not sure how to do:

Attachment 30287

Could really use a small explanation, or just the steps alone so I can interpret it myself

$2^{x+3}-2^x=224$

$2^x\cdot 2^3 - 2^x = 224$

$2^x\left(2^3-1\right)=224$

Spoiler:
$2^x\left(8-1\right)=224$

$2^x(7)=224$

$2^x=32$

$x=5$

$4\left(2^{2x}\right)+31\left(2^x\right)-8=0$

$4\left(2^x\right)^2 + 31\left(2^x\right)-8=0$

let $u=2^x$

$4u^2+31u-8=0$

$(8+u)(4u-1)=0$

$u=\{-8, \frac{1}{4}\}$

as $u=2^x>0, u=-8\mbox{ can be disregarded.}$

$2^x=\frac{1}{4}$

$x=-2$