Solving a Quadratic Equation with Exponential Variables

  • Thread starter minase
  • Start date
In summary, the conversation was about solving a homework assignment on quadratic equations. One question involving x^x=58 proved to be challenging and the speaker was unsure how to solve it. They mentioned trying trial-and-error on a calculator but were still confused. They also noted that logarithms and integrals were not covered in their class.
  • #1
minase
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0
We had a home work solving quadratic equation. I solved each question except this one. We did not learn about lograrizm ..integral staff. I had no idea how to solve this question. x^x=58. The home work is due today I am posting this forum in school. I am despirate. Help :confused:
 
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  • #2
Just forget it.
You can't express the solution of that equation in an "ordinary" manner.
Probably, it is a typo.
 
  • #3
Well..
x^x = 58
If you do trial-and-error on your calculator... you get something like x = 3.35475 (Approximation)
Then again, I have no idea why you would get such a problem.. we never even got this far in Algebra 2. :)
 

Related to Solving a Quadratic Equation with Exponential Variables

1. How do I solve x^x = 58?

The equation x^x = 58 is a transcendental equation, meaning it cannot be solved algebraically. Therefore, we need to use numerical methods to approximate the solution.

2. What numerical methods can I use to solve x^x = 58?

Some common numerical methods for solving transcendental equations include the bisection method, the Newton-Raphson method, and the secant method. These methods involve repeatedly guessing a solution and using an iterative process to refine the guess until it is close enough to the actual solution.

3. Can I use a graphing calculator to solve x^x = 58?

Yes, you can use a graphing calculator to graph the function y = x^x and visually approximate the solution. However, this method may not be as accurate as using numerical methods.

4. Are there any special values for x that satisfy x^x = 58?

Yes, the special value x = 2 satisfies x^x = 58. However, this is the only "nice" solution and all other solutions must be approximated using numerical methods.

5. Can I solve x^x = 58 using algebraic manipulation?

No, it is not possible to solve x^x = 58 using algebraic manipulation. The variable x appears in both the base and exponent of the equation, making it impossible to isolate x on one side of the equation.

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