The total number of solutions (real) of equation: $2^x+3^x+4^x-5^x=0 ?$
I have no idea how to solve this problem. Can someone point me in the right direction?
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6It is easier to look at $(2/5)^x+(3/5)^x+(4/5)^x=1$. – André Nicolas Jul 16 '16 at 10:44
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1By Laguerre's extension of Descartes' rule of sign, the number of real solution is $1$. See the statement and refs in my answer to a similar question. – achille hui Jul 16 '16 at 11:34
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Given $$2^x+3^x+4^x=5^x\Rightarrow \underbrace{\left(\frac{2}{4.5}\right)^x+\left(\frac{3}{4.5}\right)^x+\left(\frac{4}{4.5}\right)^x}_{\bf{Strictly\; decreasing \; fun.}} = \underbrace{\left(\frac{5}{4.5}\right)^x}_{\bf{Strictly\; increasing \; fun.}}$$
So these two function Intersect each other at exactly one point.
juantheron
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5Then, we could use $4.999$ instead of $4.5$. Don't you think that André Nicolas's suggestion would even be easier since lhs will be decreasing and rhs a constant ? – Claude Leibovici Jul 16 '16 at 11:32
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$$2^x+3^x+4^x=5^x$$
$${\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x} =1$$ $$y={\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x}$$ $$y=1$$ Since the function is always decreasing . It will meet $y=1$ only once .
Graph will make things more easier .
Aakash Kumar
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