Calc Real $x$ in Exponential Equation

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SUMMARY

The equation $3^x + 6^x = 4^x + 5^x$ has been analyzed, revealing that the only real solutions are $x = 0$ and $x = 1$. The function defined as $f(x) = 3^x - 4^x - 5^x + 6^x$ vanishes at these points. The derivative at $x=0$ is negative, indicating that $f(x)$ is positive for $x < 0$ and negative for $x > 1, confirming no additional roots exist beyond these values.

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  • Understanding of exponential functions and their properties
  • Knowledge of derivatives and their implications on function behavior
  • Familiarity with logarithmic functions, specifically natural logarithms
  • Basic algebraic manipulation skills
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  • Learn about the application of derivatives in finding function extrema
  • Explore logarithmic identities and their applications in solving equations
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Mathematicians, students studying calculus, and anyone interested in solving exponential equations will benefit from this discussion.

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Calculation of all real values of $x$ for which $3^x+6^x = 4^x+5^x$
 
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Re: exponential equation

jacks said:
Calculation of all real values of $x$ for which $3^x+6^x = 4^x+5^x$

[sp]Let's consider the function...

$\displaystyle f(x) = 3^{x} - 4^{x} - 5^{x} + 6^{x}\ (1)$

It is immediate to realize that f(x) vanishes in x=0 and x=1. The derivative in x=0 is...

$\displaystyle f' (0) = ln \frac{9}{10}< 0\ (2)$

... so that around x=0 is f(x)>0 for x<0 and vice versa. If x is negative then the term $3^{x}$ is dominating and is f(x)>0, so that no negative roots exist. For x>1 the term $6^{x}$ is dominating so that there is no roots greater than 1. The conclusion is that x=0 and x=1 are the only roots...[/sp]

Kind regards

$\chi$ $\sigma$
 

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