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What is

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

Guest Sep 1, 2014

Best Answer 

 #1
avatar+92781 
+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+92781 
+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

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