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April 6th, 2017, 08:20 PM   #1
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exponential equation

number of real solution of $\displaystyle 2017^x-2016^x = x$
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April 6th, 2017, 08:31 PM   #2
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$x=0,~x=1$

2 real solutions
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April 7th, 2017, 07:34 PM   #3
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Thanks romsek got it,

$\displaystyle \displaystyle \frac{2017^x-2016^x}{2017-2016} = x$

Using LMVT , we have $\displaystyle f'(t) = x, f(t) = t^x\;, t \in \left[2016,2017\right]$

So $\displaystyle xt^{x-1} = x$. So we get $\displaystyle x=0\;, 1$
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