<s>[INST] What are the number of solutions of $11^x+13^x+17^x-19^x$? 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??
 [/INST] 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. 
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