My attempt: $ 11^x+13^x+17^x-19^x=0\implies (\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x=1 $.
Now taking limit as $x\rightarrow\infty$,we get $0=1$,which is absurd.Hence,the equation has no solution.
Is it correct??
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there is a real solution – Dr. Sonnhard Graubner Aug 28 '17 at 10:37
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$11^x+13^x+17^x-19^x$ is positive for $x=0$ and negative for large $x$ and is continuous so there is a positive solution – Henry Aug 28 '17 at 10:37
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a numerical method is a good choice – Dr. Sonnhard Graubner Aug 28 '17 at 10:42
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No, it is not correct. The limit only shows that there are no solutions for $x$ sufficiently large. – Michael Burr Aug 28 '17 at 10:50
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Why does the limit mean that there is no solution? All it really shows is that there aren't infinitely large solutions... – Thomas Andrews Aug 28 '17 at 16:45
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Let $f(x)= 11^x+13^x+17^x-19^x=19^x[(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1]$.
Clearly,$g(x)=(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1$ is a strictly decreasing function.Also,$\lim_{x\to \infty}g(x)=-1 $ & $\lim_{x\to -\infty}g(x)=\infty$.Thus $f(x)=0$ has exactly one solution.
Styles
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As you have been said
$$ \lim_{x \rightarrow \infty} \Big( (\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x \Big) =0; $$
this means that there is an upper bound
for the set of solutions to the eqation
$(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x=1 $;
you can take $5$ as an upper bound, but this is not sharp.
Davood
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