How to Find z in a Complex 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!

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.