If , and , what is the product of the possible values of x?
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Wait, is p = 1/32^5 or (1/32)^5? The formatting is ambiguous and it completely changes the value of p.
It's 1/(32^5), since 32^5 is in the denominator. So p = 32^(-5).
Yeah that's how I read it too, p = 2^(-25).
Since bases are equal, x^p = p^(1/x). At this point I rewrote p = 2^(-25) and tried x = 2^k. Then x^p = 2^(k * 2^-25) and p^(1/x) = 2^(-25/2^k). So k/2^25 = -25/2^k, i.e. k * 2^k = -25 * 2^25. Now k = -25 works, but is there another solution? The function k*2^k is not one-to-one on negatives...
k*2^k for real k is actually strictly increasing? derivative is 2^k(1 + k ln2), which is negative when k < -1/ln2 ≈ -1.44, so it decreases then increases. So there could be two solutions.
Got k = -25 then checked k = -? I ended up with k*2^k = -25*2^25. Obvious k=-25, then product of x's would be 2^(sum of k's). If the other k isn't nice I don't see how to get 27 or 54.
Maybe use Lambert W? But this is supposed to be solvable by hand.
This is way too hard for me. I got p = 2^-25 and then got stuck. Respect to anyone who solves this in 2 minutes.
What it tests
Your grasp of exponent rules — multiplying and dividing powers, power-of-a-power, and negative/zero exponents.
Common trap
Adding exponents when multiplying bases that have the same exponent, or confusing (x^a)^b with x^(a·b).