Junior Math Olympiad Algebra Problem

Junior Math Olympiad Algebra Problem

[XJWill]

I used a slight variation on that method:
72^x * 48^y = 6^(x*y)
2^x * 6^(2*x) * 6^y * 2^(3*y) = 6^(x*y)
2^(x+3*y) * 6^(2*x + y) = 6^(x*y), therefore:
x+3*y = 0 and 2*x + y = x*y
x = -3*y and 2*(-3*y) + y = -3*y*y
5*y = 3*y^2

[sidharthiyer]

I also started this question with the same approach as yours (comparing powers both sides), but instead of subtracting equations directly, I divided all terms by xy both sides.

Which gave me 3/y + 4/x = 1 and 2/y + 1/x = 1.
I assumed 1/y as p and 1/x as q, which gave me equations 3p + 4q = 1 and 2p + q = 1 respectively. It gave solutions easily.

Keep up the good work, Guruji. 🙇‍♂️🙇‍♂️🙇‍♂️ Your videos are awesome.

[manoranjansarker]

excellent math problem.salute from🇧🇩🇧🇩

[rayjify]

It may have many more sets of solutions.

[MichaelRothwell]

As suggested by rayj88ify, if you remove the conditions that x and y be nonzero rational numbers you obtain many more sets of solutions.
For any given real value of x we have
72ˣ 48ʸ = 6ˣʸ
Take logs base 6:
log₆ 72ˣ + log₆ 48ʸ = xy
xlog₆72 + ylog₆ 48 = xy
xlog₆(2×6²) + ylog₆(2³×6) = xy
x(log₆2 + 2) + y(3log₆2 + 1) = xy
x(log₆2 + 2) = y(x - 3log₆2 - 1)
y = x(log₆2 + 2)/(x - 3log₆2 - 1), for any given real x.

Note that for a rational solution, as log₆2 is irrational, the ratio of the coefficient of log₆2 in the numerator to the coefficient of log₆2 in the denominator must be the same as the ratio of the remaining terms in the numerator to the remaining terms in the denominator, i.e. 1/-3 =2/(x-1), so x-1=-6, x=-5, and y=x(1/-3)=5/3, as before.

[田村博志-zy]

Let a, b be real numbers satisfying
a, b > 0,
2^a = 3^b.
Then a/b is irrational. This is the key in this problem.

I recommend that we prove this. Very easy.