MHB Calc Real $x$ in Exponential Equation

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The equation \(3^x + 6^x = 4^x + 5^x\) is analyzed through the function \(f(x) = 3^x - 4^x - 5^x + 6^x\). It is established that \(f(x)\) equals zero at \(x=0\) and \(x=1\). The derivative at \(x=0\) indicates that \(f(x)\) is positive for \(x<0\) and negative for \(x>0\). For \(x>1\), the term \(6^x\) dominates, confirming no roots exist beyond this point. Thus, the only real solutions are \(x=0\) and \(x=1\).
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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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