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I was looking at this problem https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_9

Let $x,$ $y,$ and $z$ be positive real numbers that satisfy $2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.$The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

The solution begins like this:

Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't $0$, of course), so to simplify the problem let us assume without loss of generality that$2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.$

I am wondering why we can assume that the expression would be equal to a certain value because the (two) equations do not imply that $2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2$, although the equations themselves are satisfied when the expression equals $2$.

Why is this a sufficient answer (i.e. could there be a different value of $xy^5z$ and/or $p+q$ if the expressions don't evaluate to $2$?) Also, does this work in the general case i.e. given three expressions from variables $x, y, z$ which are equal to each other, they can equal any value, so can we just assume they are equal to a particular constant to solve for an expression involving $x, y, z$?

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This is NOT a full solution. Since contestants are asked to calculate the value of $p+q$ (without proof), they only need to obtain a numerical answer. I like to call this method "obtain answer by special case", where you can "tweak" the question to make it easier to calculate, as long as it still fits the constraints. If the question does indeed have a fixed solution, we will obtain the correct answer from our special case.

Since time is of the essence in competitions, this is used whenever possible (e.g. turning quadrilaterals into squares, letting constants be zero, etc.) I'd imagine the full solution to be quite a bit more complicated.

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  • $\begingroup$ Are you saying that because there should be only one solution, if you find one solution that works using the "backward-solving" technique, then that would be the correct answer? $\endgroup$
    – pblpbl
    Sep 27 '20 at 16:38
  • $\begingroup$ Yes. But I cannot stress enough that this technique cannot yield the full solution. $\endgroup$
    – player3236
    Sep 27 '20 at 16:39
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If you want the long way, write $$2\log_{x}(2y) =k\implies \frac{2 \log (2 y)}{\log (x)}=k\implies y=\frac{1}{2}x^{k/2}\tag1$$ $$2\log_{2x}(4z) =k\implies\frac{2 \log (4 z)}{\log (2 x)}=k\implies z=2^{\frac{k}{2}-2} x^{k/2}\tag2$$ $$\log_{2x^4}(8yz)=k\implies\frac{\log (8 y z)}{\log \left(2 x^4\right)}=k\tag 3$$

Using the results given by $(1)$ and $(2)$, we have now $$\frac{\log \left(2^{k/2} x^k\right)}{\log \left(2 x^4\right)}=k\implies x= {2^{-\frac 16}} \quad y=2^{-\frac{k}{12}-1}\quad z=2^{\frac{5 k}{12}-2}\tag 4$$ Then $$xy^5z=\frac 1 {2^{7+\frac 16}}=\frac 1 {2^{\frac{43}{6}}}\implies p=43 \quad q=6\implies p+q=49$$

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If you have one more unknown than independent equations, you usually assume one unknown as a parameter. So you can determinate different values of the remaining unknowns as the parameter varies.

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