Prove that $10^x+11^x+12^x=13^x+14^x$ has an unique solution over $\mathbb R$. By inspection the equation is true for $x=2$
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wolframalpha's hint: http://www.wolframalpha.com/input/?i=14^x%2B13^x+-+12^x-+11^x+-+10^x – ninjaaa Oct 22 '15 at 08:04
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The equation is equivalent to $$\left(\frac{10}{13}\right)^x+\left(\frac{11}{13}\right)^x+\left(\frac{12}{13}\right)^x=1+\left(\frac{14}{13}\right)^x.$$ It is clear that there are no negative solutions. And for positive $x$, the left side is decreasing, and the right side is increasing, so there is at most one value of $x$ where they are equal.
André Nicolas
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+1 Clear answer explains why it should be unique, if a solution exists. And no algebra/number theory has been used. – P Vanchinathan Oct 22 '15 at 08:19