Pre-Calculus help?!?!
Favorites|Homepage
Subscriptions | sitemap
HOME > > Pre-Calculus help?!?!

Pre-Calculus help?!?!

[From: ] [author: ] [Date: 12-11-19] [Hit: ]
The answer is -1 and 0,2.The answer is 180.384, if that helps. I just cant figure out what work to show.......
Solve. The e isn't a variable, we are doing logarithms.
1. 2x^2e^(2x)+2xe^(2x)
The answer is -1 and 0, but i just cant figure out how to get that
2. log8x-log(1+x^(1/2))=2
The answer is 180.384, if that helps. I just cant figure out what work to show.

Thanks in advance for any help you can offer!!

-
1) Assuming 2x^2e^(2x)+2xe^(2x) = 0
2xe^(2x)(x+1) = 0
x = 0 or -1

2) log [8x/(1+x^(1/2))] = log 100
[8x/(1+x^(1/2))] = 100
8x-100sqrt(x)-100 = 0
2x-25sqrt(x)-25 = 0
Use the quadratic formula
sqrt(x) = 13.4307
x = 180.384
1
keywords: Pre,help,Calculus,Pre-Calculus help?!?!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .