Solving Equation for Lambda in Terms of A

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Homework Help Overview

The discussion revolves around solving for the variable lambda in terms of A, particularly focusing on handling exponential terms and logarithmic properties in the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore methods such as taking the logarithm of both sides and reasoning about the equality of bases in exponential functions. There are questions regarding the cancellation of lambda terms and the implications of such cancellations in the context of the problem.

Discussion Status

Some participants have offered guidance on using logarithmic properties and expanding binomials, while others are questioning the validity of cancelling terms. The discussion reflects a mix of interpretations and approaches without reaching a consensus.

Contextual Notes

Participants are navigating through the complexities of exponential equations and logarithmic identities, with some expressing uncertainty about the implications of their manipulations on the variable lambda.

ACLerok
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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
 
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You can take the log of both sides
 
Or, you can use logical reasoning to conclude that if the bases on each side are equal, then...?
 
"the bases on each side are equal" and exponential is a one-to-one function!
 
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ACLerok said:
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.
 
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