+0  
 
0
267
1
avatar

What is

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

Guest Sep 1, 2014

Best Answer 

 #1
avatar+91458 
+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
Sort: 

1+0 Answers

 #1
avatar+91458 
+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

12 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details