# Thread: Help with an equation.

1. Hi

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

2. 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

3. 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.

4. 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.

5. 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...

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