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发帖数: 244 | 1 We need to solve $x^12=2^x$, first,
$x\neq 0$, when $x>0$, we rewrite it
as $f(y)=y^y=2^{-\frac{1}{12}}$, where
$y=\frac{1}{x}$, $y^y$ has only critical point
at $y=e^{-1}$, and we have $f(0+)=1$,
$f(+\infty)=\infty$.
Notice $0<2^{-\frac{1}{12}}<1$. So since
$f(e^{-1})<2^{-\frac{1}{12}}$,
we have 2 solution in this case.
Next suppose $x<0$, let $y=-\frac{1}{x}$, we
have $f(y)=y^y=2^{\frac{1}{12}}$, notice
$2^{\frac{1}{12}}>1$, we have
1 solution in this case.
Draw the graph of $f(y)=y^y$$, you | d*z
发帖数: 150 | 2 The result is 3.
We need to find roots for 2^x-x^12=0
That's
x log2 - 12 logx=0
Because
d(xlog2-12logx)=log2-12/x
So that xlog2-12logx increasing while x<0 and first increase then decrease for
x>0.
So that's at most 3 roots for the equation.
2^(-Inf)-(-Inf)^12<0
2^0-0^12>0
So there's one root for x<0.
2^0-0^12>0
2^(2)-2^12 <0
2^(Inf)-Inf^12>0
So there's two roots for x>0
Total there're 3 roots. |
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