Determine the total number of ordered pairs(x,y) of positive integers with 2≤ x,y ≤ 2023 that satisfy this equation:
xlogy(x-4) = ylogx(yx-3)
I cannot find any solutions in the required range/domain
Unless I made an error:
The equation ultimately simplifies to
-4*(lnx)^3 = (lny-3lnx) * (lny)^2
Substitute s = lnx and t = lny
then substitute r = t/s
r^3 - 3r^2 + 4 = 0
r^2(r-3) + 4 = 0 from which
the domain is {r < 3}
r = -1 is a solution
r^3 - 3r^2 + 4 = (r+1)(r^2 - 4r - 4)
r = {-1, 2-2√2} reject 2+2√2 because > 3
lny/lnx = {-1, 2-2√2}
case 1: lny = -lnx; lny = ln(1/x); y = 1/x; xy=1
(x,y) = (1,1) would be a solution but 2 ≤ x,y ≤ 2023 so it's not
case 2: lny/lnx = 2-2√2
lny = 2lnx-2√2lnx
lny = ln(x^2) - lnx(x^2√2)
lny = ln ((x^2)/(x^2√2))
lny = ln (x^(2-2√2))
y = x^(2-2√2)
which passes through (1,1) but then if y>x then x<1 and if y<x then y<1 and (1,1) is excluded by the given conditions.
Edited on July 30, 2023, 8:15 am
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Posted by Larry
on 2023-07-30 08:13:50 |
I find ...
