Hard Algebra 2 Problem?

[de.bug]

x = 3^6 * 2^12
y = 3^8 * 2^8
x^x * y^y = z^z for some integer z

Find z.

I took the log of both sides, but don't know what to do next (I'm not even sure taking log of both sides will produce anything).

[I like Serena]

x = 3^6 * 2^12
y = 3^8 * 2^8
x^x * y^y = z^z for some integer z

Find z.

I took the log of both sides, but don't know what to do next (I'm not even sure taking log of both sides will produce anything).

Welcome to PF, de.bug! :smile:

What about simply substituting x and y in x^x * y^y and simplifying?

[D H]

What about simply substituting x and y in x^x * y^y and simplifying?
That is a going to get more than a bit unwieldy. x and y are already fairly big numbers. x^x and y^y are incredibly large numbers.


I took the log of both sides, but don't know what to do next (I'm not even sure taking log of both sides will produce anything).

Hint: You need to make some assumption about the nature of z. What do the nature of x and y suggest? With the right assumption, taking the log of both sides will lead to the solution.

[I like Serena]

That is a going to get more than a bit unwieldy. x and y are already fairly big numbers. x^x and y^y are incredibly large numbers.

It won't be unwieldy.
As long as you don't actually calculate anything, but stick to powers of 2 and 3, the result is obtained in 6 lines.