+0  
 
0
461
1
avatar

What is

1/2x^(-1/2) *((2xe^x)/ln(3x^2 +2))

Guest Sep 1, 2014

Best Answer 

 #1
avatar+93665 
+5

$$1/2x^{-1/2} *((2xe^x)/ln(3x^2 +2))\\\\
=\frac{x^{-1/2}}{\not{2}}*\frac{\not{2}xe^x}{ln(3x^2 +2)}\\\\
=\frac{1}{1}*\frac{x^{(1-1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{x^{(1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{\sqrt{x}e^x}{ln(3x^2 +2)}\\\\$$

 

I don't think that there is anything else to be done with this.  

 

Has anyone got any further/different thoughts?

Melody  Sep 1, 2014
 #1
avatar+93665 
+5
Best Answer

$$1/2x^{-1/2} *((2xe^x)/ln(3x^2 +2))\\\\
=\frac{x^{-1/2}}{\not{2}}*\frac{\not{2}xe^x}{ln(3x^2 +2)}\\\\
=\frac{1}{1}*\frac{x^{(1-1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{x^{(1/2)}e^x}{ln(3x^2 +2)}\\\\
=\frac{\sqrt{x}e^x}{ln(3x^2 +2)}\\\\$$

 

I don't think that there is anything else to be done with this.  

 

Has anyone got any further/different thoughts?

Melody  Sep 1, 2014

17 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.