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The problem is : To find number of real solutions of $ \ \ 2011^x$ $+$ $2012^x$ $+$ $2013^x$ $=$ $2014^x$

My attempt : I first tried to see if the equation has zero solutions ;i.e ,the LHS of the equation was even and so was the RHS so I couldn't prove it had zero solutions.

Next I tried finding the derivative of $ \ \ 2011^x$ $+$ $2012^x$ $+$ $2013^x$ $-$ $2014^x \ $ to get an idea of the graph and it turned out to be : $$2011^x \ln 2011 +2012^x \ln 2012 +2013^x \ln 2013 - 2014^x \ln 2014 $$

Now I am stuck ,how do I proceed

Albus Dumbledore
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2 Answers2

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Divide by $2014^x$ to get $$a^x+b^x+c^x=1$$ The derivative is $(\ln a) a^x+(\ln b)b^x+(\ln c)c^x<0$ yet at $x=0$, the LHS $=3>1$ and at $x\to\infty$, LHS$\to0$. So only one solution.

Chrystomath
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  • Since $a, b, c \in (0, 1)$ then the function $ g (x) = a ^ x + b ^ x + c ^ x -1 $ is strictly decreasing and consequently the square has at most one solution. As in $R$ the continuous function $g$ changes its sign, it results that the equation has a unique solution. (Intermediate value theorem). – medicu Jul 29 '20 at 15:24
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Define $f(x)=2011^x+2012^x+2013^x-2014^x$ and $g(x)=3\cdot 2013^x-2014^x$ then you usee that $g(0)=f(0)=2$ and $g(x)>f(x)$ for $x>0$.

Assume $g(a)=0$ that is $3\cdot 2013^a-2014^a=0$ with $a=\frac{\ln 3}{\ln 2014-\ln 2013}$ as a root to $g$.

Use $f(0)=g(0)=2$ and $f(x)<g(x)$ for $x>0$ and $g(a)=0$ to conclude that $f(x)$ has a root in the interval $(0,a]$. Try use interval halving to find that root.

Mikael Helin
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