Getting rid of multiple e's in equation e^a - e^b = 2e^c

[diargos]

My friend asked me if it is possible to "remove the base e" from the equation in the form of:

e^a - e^b = 2e^c

My first thought was to take the log of each side, to work towards getting the equation in the form of something like log(a) - log(b) = 2*log(c). But obviously I can't just do that, since it's log(e^a - e^b). What can I do with that whole term? Or what other approach should I look at for solving this problem?

Thanks!

 

[topsquark]

My friend asked me if it is possible to "remove the base e" from the equation in the form of:

e^a - e^b = 2e^c

My first thought was to take the log of each side, to work towards getting the equation in the form of something like log(a) - log(b) = 2*log(c). But obviously I can't just do that, since it's log(e^a - e^b). What can I do with that whole term? Or what other approach should I look at for solving this problem?

Thanks!

If by removing the e's you mean something like a - b = 2c or something? If that's the case then there is no way to do it. The best idea I can suggest is to let m = ln(a), b = ln(n) , and c = ln(p) which would give you m - n = 2p. This may or may not seem like we've done it but note that if we solve for m, for example, m = n + 2p we find we can't back substitute to find the vale of a without having the ln terms there, which is just as bad.

-Dan

 

[mmm4444bot]

My friend asked me if it is possible to "remove the base e" from [this] equation …

e^a - e^b = 2e^c

… [what] approach should I look at for [answering this question for my friend]?

You should ask your friend why they're asking. If we remove the base, there is no equation (it's like asking whether it's possible to remove the decimal points from one's bank statement; if we do that, it's no longer our bank statement).

I'll guess your friend wants to find solutions for {a,b,c} that satisfy the given equation. If this is correct, then tell your friend they don't have enough information to find formulas (i.e., symbolic solutions). We need more information about how a, b and c are related. Using algebraic methods, we generally need N non-equivalent equations (relationships) when we're trying to find N variables. Your friend is trying to find three variables, but has only one equation.

We can find some solutions, however. Let's assume your friend decides that a-b=2c is also true. If so, then your friend could choose one value (like c=-1 or c=2, for example) and then express the others (a and b, in this example) in terms of the choice for c. Maybe the choice leads to a solution; maybe not.

Let's see what MVR5 says.

When c = -1, then a and b are not Real numbers

When c = 2, then a ≈ 2.711632627 and b ≈ -1.288367373

When c = \(\displaystyle e\), then a ≈ 3.415792938 and b ≈ -2.020770719 :cool:

 

[mmm4444bot]

After using scratch paper, I see that it's straightforward to solve for c in terms of a and b, using logarithms.

In other words, without assuming a-b=2c, your friend can get a formula for c, instead of guessing followed by hoping there's corresponding, Real solutions for a and b.

So, they can pick values for a and b, instead of c, to approximate a solution.

When a = 2 and b = -1, then c ≈ 0.8481776744

These solutions came from numerical methods (aka: the brute-force approach). :cool:

 

[Steven G]

My friend asked me if it is possible to "remove the base e" from the equation in the form of:

e^a - e^b = 2e^c

My first thought was to take the log of each side, to work towards getting the equation in the form of something like log(a) - log(b) = 2*log(c).

I see how you went from e^a - e^b to log(a) - log(b). You used the UNTRUE fact that log(x-y) = log x - logy. You even said it was wrong (good!). I am more concerned of the rhs. log(2e^c) is not 2*log(c). log(2e^c) = log(2) + c. I am using log to represent the natural log.