$$g(x) = \frac{100}{1+2^{-x}}$$
Ok, i have this expression and my task is to find the inverse. My answer to that is -ln2((100-x)/x). Which is wrong when i test it. Can someone help me with this?
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Why did you unaccept my answer in favor of an answer that was published later??? – barak manos Jan 25 '15 at 21:03
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im sorry. it was marked unaccepted so i think something is bugged here :P – kirkegaard Jan 25 '15 at 21:28
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You cannot accept two answers. Once you accept a second answer, the first one is automatically unaccepted. – barak manos Jan 25 '15 at 22:03
3 Answers
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$g(x)=\frac{100}{1+2^{-x}}\implies$
$g(x)\cdot(1+2^{-x})=100\implies$
$1+2^{-x}=\frac{100}{g(x)}\implies$
$2^{-x}=\frac{100}{g(x)}-1\implies$
$-x=\log_2(\frac{100}{g(x)}-1)\implies$
$x=-\log_2(\frac{100}{g(x)}-1)$
barak manos
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$$g(x) = \frac{100}{1+2^{-x}}$$
$$\frac{100}{g(x)} = 1+2^{-x}$$
$$\frac{100}{g(x)} -1 = 2^{-x}$$
$$\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = -x$$
$$-\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = x$$
Loreno Heer
- 4,460
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I can let $y=\frac{100}{1+2^{-x}}$.
$$ 1+2^{-x} = \frac{100}{y}$$ $$ 2^{-x} = \frac{100}{y}-1$$ $$ 2^{-x} = \frac{100-y}{y}$$ $$2^x = \frac{y}{100-y}$$ $$x = \log_2 \left ( \frac{y}{100-y}\right)$$
cgo
- 1,810