# Anyone help me to solve this algebra....

• Dec 9th 2012, 02:30 AM
sharmala
Anyone help me to solve this algebra....
X^5 e^-3 In x + 4x =21 help me to solve this..thank u
• Dec 9th 2012, 02:45 AM
MarkFL
Re: Anyone help me to solve this algebra....
Do you mean:

x^5·e^(-3·ln(x)) + 4x = 21 ?

Do you see how the use of bracketing symbols makes the equation clear? Also, you should not mix uppercase and lowercase variable names, as these technically refer to different variables. Lastly, the natural log function is expressed as "ln" not "In."

Now, given that I have interpreted correctly this time, do you have any thoughts on what you should do first?
• Dec 9th 2012, 03:15 AM
rcs
Re: Anyone help me to solve this algebra....
Quote:

Originally Posted by MarkFL2
Do you mean:

x^5·e^(-3·ln(x)) + 4x = 21 ?

Do you see how the use of bracketing symbols makes the equation clear? Also, you should not mix uppercase and lowercase variable names, as these technically refer to different variables. Lastly, the natural log function is expressed as "ln" not "In."

Now, given that I have interpreted correctly this time, do you have any thoughts on what you should do first?

Hope this is right

x^5 (e^(-3lnx)) + 4x =21
x^5 (e^(ln x^-3))+4x = 21
x^5 (x^-3) +4x =21
x^5/x^3 + 4x - 21 = 0
x^2 + 4x -21 = 0
(x + 7) (x-3) = 0

x = -7 Or x = 3
• Dec 9th 2012, 03:23 AM
MarkFL
Re: Anyone help me to solve this algebra....
What happens when you substitute the two roots you have found into the original equation?
• Dec 9th 2012, 09:59 PM
rcs
Re: Anyone help me to solve this algebra....
Kindly substitute Sir MarkFL2. ... I think it would be best if it is you
• Dec 9th 2012, 10:05 PM
MarkFL
Re: Anyone help me to solve this algebra....
Actually, I think you would learn more if you do it...and it is not difficult, but you will find a problem with one of the roots.