How to Find z in a Complex Algebra 2 Problem?

In summary, the problem involves finding an integer z that satisfies the equation x^x * y^y = z^z, where x = 3^6 * 2^12 and y = 3^8 * 2^8. By taking the log of both sides and making an appropriate assumption about the nature of z, the solution can be obtained in 6 lines by substituting x and y into x^x * y^y and simplifying. This method avoids any calculations involving large numbers.
  • #1
de.bug
2
0
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).
 
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  • #2
de.bug said:
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?
 
  • #3
I like Serena said:
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.


de.bug said:
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.
 
  • #4
D H said:
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.
 

Related to How to Find z in a Complex Algebra 2 Problem?

What is the purpose of solving hard algebra 2 problems?

The purpose of solving hard algebra 2 problems is to deepen your understanding of advanced algebra concepts and strengthen your problem-solving skills. It also prepares you for more difficult math courses and real-world applications where algebra is used.

Why are algebra 2 problems considered difficult?

Algebra 2 problems are considered difficult because they require a combination of complex algebraic concepts, critical thinking, and multiple steps to solve. These problems often involve equations with multiple variables and unknowns, making them challenging to solve.

What are some strategies for approaching hard algebra 2 problems?

Some strategies for approaching hard algebra 2 problems include breaking the problem down into smaller, more manageable steps, using algebraic properties and formulas, and checking your work for errors. It's also helpful to practice regularly and seek help from a teacher or tutor if needed.

How can I improve my skills in solving hard algebra 2 problems?

To improve your skills in solving hard algebra 2 problems, it is important to review and practice fundamental algebra concepts, such as factoring, solving equations, and working with exponents and radicals. It's also beneficial to challenge yourself with increasingly difficult problems and seek help and feedback from peers or a teacher.

What are some real-world applications of solving hard algebra 2 problems?

Solving hard algebra 2 problems has many real-world applications, such as in engineering, physics, economics, and computer science. For example, algebraic equations are used to model and solve problems in structural engineering and to analyze data in economics and finance. Algebra 2 concepts are also applied in computer programming to write algorithms and solve complex problems.

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