Help with an equation.

**JBroeng** · May 4th 2010, 12:39 AM

Hi

Im Have this equation and I have the result but I can´t get from a to b....

**running-gag** · May 4th 2010, 10:31 AM

Hi

There are several errors in your calculations

At the end of the 1st page there is a + instead of a - inside the square root. So is it in the 2 first equations in the second page

Then you made an error in the second equation of the second page in the simplification by $\alpha$

The right equation is $\beta e^{-\alpha r^V} = \frac{1+\sqrt{1+k_2 \alpha^2}}{k_1 \alpha}$

But I agree with you : the disappearance of $\beta$ is strange

**JBroeng** · May 4th 2010, 11:47 PM

Originally Posted by running-gag

Hi

There are several errors in your calculations

At the end of the 1st page there is a + instead of a - inside the square root. So is it in the 2 first equations in the second page

Then you made an error in the second equation of the second page in the simplification by $\alpha$

The right equation is $\beta e^{-\alpha r^V} = \frac{1+\sqrt{1+k_2 \alpha^2}}{k_1 \alpha}$

But I agree with you : the disappearance of $\beta$ is strange

Thank you SOOOOOO MUUUUCH

But what will you get r^v equal to??
Will you devide the right side first with beta and then isolate r^v?

Thank you for your help.

**HallsofIvy** · May 5th 2010, 01:51 AM

Yes, once you have $e^{-\alpha r^V}= \frac{1+\sqrt{1+ k_2\alpha^2}}{\beta k_1\alpha}$ , take the logarithm of both sides.

**JBroeng** · May 5th 2010, 04:01 AM

Originally Posted by HallsofIvy

Yes, once you have $e^{-\alpha r^V}= \frac{1+\sqrt{1+ k_2\alpha^2}}{\beta k_1\alpha}$ , take the logarithm of both sides.

No You cant just do that....it is the whole fraction you have to divide with beta...

**JBroeng** · May 5th 2010, 10:11 AM

Never mind the last message ....Have found out that you are right. Sorry and thanks....

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