Thread: Anyone help me to solve this algebra....

X^5 e^-3 In x + 4x =21 help me to solve this..thank u

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?

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

What happens when you substitute the two roots you have found into the original equation?

Kindly substitute Sir MarkFL2. ... I think it would be best if it is you

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.