Solving Equation for Lambda in Terms of A

[ACLerok]

I am trying to solve for the variable lambda in terms of A. After multiplying the denominator term over to the other side, how do I go on from there? I don't know how to get rid of the exponential terms.

Thanks on advance.

http://img521.imageshack.us/img521/4006/picture1ug1.th.png

 

[klile82]

You can take the log of both sides

 

[atqamar]

Or, you can use logical reasoning to conclude that if the bases on each side are equal, then...?

 

[HallsofIvy]

"the bases on each side are equal" and exponential is a one-to-one function!

 

[ACLerok]

After cancelling the ln's I end up with the equation below, but then it seems the lambdas cancel each other out. Is that correct? apparently from the solutions, lambda does not cancel.
http://img516.imageshack.us/img516/5822/picture1nd7.th.png

 

[stewartcs]

After cancelling the ln's I end up with the equation below, but then it seems the lambdas cancel each other out. Is that correct? apparently from the solutions, lambda does not cancel.
http://img516.imageshack.us/img516/5822/picture1nd7.th.png

You're equation looks correct so far, but you can reduce it to lamba in terms of A.

Hint: Expand the binomial on the RHS.