The problem is to solve $3^x+4^x+14^x=10^x+11^x$ for $x>0$.
I have noticed that $x=1$ and $x=2$ are solutions; probably they are the only solutions, but I do not know how to prove it; I have tried some monotony and convexity arguments.
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Is x an integer? – Upax Dec 29 '19 at 16:03
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No; it is positive real. – JohnnyC Dec 29 '19 at 16:05
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2Prove $14^x>10^x+11^x$ for all $x$ exceeding some $x_0$, then at least you have a restriction on the size of solutions. – Gerry Myerson Dec 29 '19 at 16:06
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1@Gerry Myerson: Yes, indeed; one obtains that any solution of the equation is less than 2.42... – JohnnyC Dec 30 '19 at 09:40
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OK, so, at least we've ruled out any further integer solutions. – Gerry Myerson Dec 30 '19 at 15:01
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8Does this answer your question? An exponential equation over positive real numbers – LHF Feb 06 '20 at 08:57