- Thread starter akunzler
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You have not posed a question. You must tell us exactly what is needed.How do you take the inverse of \(\displaystyle y=2^x+3^x\)

Or how do you solve for \(\displaystyle y\) in \(\displaystyle x = 2^y + 3^y\)

I and all the math teachers at my high school are stumped.

What is the object of the quest? Look at the graphs.

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But not every problem can be solved algebraically! There may be some trick I'm not thinking of, but it wouldn't surprise me at all if this can't be done.

Do you have any reason to think it should?

I am not actually sure it can be done, but I figured this would be an okay place to ask

I am trying to find a function to describe a set of points, but the function I found (through a mixture of experimentation and research) was in terms of y. I want to be able to input x and get y, so I needed to solve it for y (effectively taking the inverse if the variables were swapped)

The equation was pretty complicated, plotting \(\displaystyle \left(\frac{3\cdot2^{y_{1}+1}Z+\left(-2\right)^{y_{1}}+\left(-1\right)^{2y_{1}+1}}{3},y_{1}\right)\) with \(\displaystyle Z\) being all integers plotted the data set I was given.

I don't know if I can take the inverse or manipulate this to get it in terms of x, or if I have to come at the data from a different direction. Starting with a more simple equation based on similar concepts: \(\displaystyle y=2^x+3^x\) seemed a good place to start.

I did know that I was asking the same question two different ways, but I didn't want to ask for the inverse of \(\displaystyle y=2^x+3^x\) and be told it was \(\displaystyle x=2^y+3^y\), because I specifically need to solve for y.

P.S. I asked more specifically about this data set here, but I haven't found an answer yet.

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But, as I said, I doubt that there is an

My point was that many students, having only been given problems in class that can be solved by the methods they are learning, assume that an inverse can be found for any function. Your teachers should be aware of this, but (like me, actually) don't have an answer but don't want to definitively state that it is utterly impossible. Sometimes, as I tell my students, it's not your fault that a problem has no solution, but the problem's! The trick is to know the difference.

For some confirmation that it can't be done, see

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For example if

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Would you like some extra hints so that you can find it too? (I'd rather not give you the answer directly because you won't learn as much from that)

NB: I don't think there is an inverse if y is not constrained in some way. (But obviously it could be done numerically.)

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I made an approximation to the equation \(\displaystyle x=2^y+3^y\). Assume a large value of y and which term dominates? This approximate equation can then easily be inverted algebraically. At this point the value obtained for y is a real number that is bigger than the desired integer value - but something simple can be done to obtain the required integer value. Then, to be complete. you can check the range of values for which the method holds. It might be a bigger range than expected.

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Rounding an approximate value can be a valid way to solve such a problem. But to make this a full solution to the problem, you need to prove conditions under which rounding is guaranteed to produce the correct value of y; and I would also want to be able to know whether a given x is actually in theI made an approximation to the equation \(\displaystyle x=2^y+3^y\). Assume a large value of y and which term dominates? This approximate equation can then easily be inverted algebraically. At this point the value obtained for y is a real number that is bigger than the desired integer value - but something simple can be done to obtain the required integer value. Then, to be complete. you can check the range of values for which the method holds. It might be a bigger range than expected.

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I was actually thinking of the simple approximation \(\displaystyle x≈3^y\)That makes sense

For example, one simple though not very accurate approximation I made was \(\displaystyle 5\log\left(\sqrt{\frac{x}{2}}\right)=y\)

which leads to \(\displaystyle y≈\frac{\log{\left(x\right)}}{\log{\left(3\right)}} \)

Then to restore the integer constraint, \(\displaystyle y=\left\lfloor \frac{\log{\left(x\right)}}{\log{\left(3\right)}} \right\rfloor\)

To check the range of values for which the method holds, I

I've tested it numerically from y=1 to 1000, and it gives the correct answer (obviously x gets very large so you need an arbitrary precision calculator). I also did a successful test when y=100,000 but I won't test any higher numbers because my computer was starting to melt at that point

I'll look up the meaning of domain, that Dr. Peterson kindly points out. I hope this doesn't scupper the idea!

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For the equation

- (constrained) domain (

- range (

The inverse function

- (constrained) domain (

- range (

The only way to be sure that an

- Use given
*x*to calculate*y*using**B** - Plug it back into
**A**, \(\displaystyle x_2=2^y+3^y\) - Check that
*x*is the same value as \(\displaystyle x_2\)