How can I solve more complex exponential equations?

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SUMMARY

The discussion focuses on solving complex exponential equations, specifically the equation 5^(x - 2) + 8^(x) = 200. Participants agree that this equation cannot be solved algebraically and recommend using numeric root-finding techniques. The Newton-Raphson method is highlighted as an effective approach for approximating the solution, yielding x ≈ 2.5421632382360203811. This method allows for achieving the desired precision in the solution.

PREREQUISITES
  • Understanding of exponential equations and their properties
  • Familiarity with numeric root-finding techniques
  • Knowledge of the Newton-Raphson method
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Newton-Raphson method in detail for numeric approximations
  • Explore other numeric root-finding techniques such as the Bisection method
  • Learn about the graphical interpretation of exponential equations
  • Investigate software tools for solving complex equations, such as MATLAB or Python's SciPy library
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in solving complex exponential equations using numeric methods.

mathdad
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I can solve equations like 4^(x) = 16 or
5^(x + 1) = 25. However, there are exponential equations that a bit more involved. The following equation has two exponentials on the left side.

Solve for x.

5^(x - 2) + 8^(x) = 200
 
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RTCNTC said:
I can solve equations like 4^(x) = 16 or
5^(x + 1) = 25. However, there are exponential equations that a bit more involved. The following equation has two exponentials on the left side.

Solve for x.

5^(x - 2) + 8^(x) = 200

I don't believe you can solve that algebraically...I would use a numeric root-finding technique, such as the Newton-Raphson method, to approximate the solution to the desired number of decimal places:

$$x\approx2.5421632382360203811$$
 
MarkFL said:
I don't believe you can solve that algebraically...I would use a numeric root-finding technique, such as the Newton-Raphson method, to approximate the solution to the desired number of decimal places:

$$x\approx2.5421632382360203811$$

Ok. Good to know. I don't feel so bad now.
 

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