The directions say "find a formula for f^-1 and verify that (f of f^-1)(x)=(f^1 of f)(x)=x. The problem is f(x)=100/(1+2^-x). Can anybody help me??
My preferred method is to set f(x) = y and solve for x in terms of y:
$\displaystyle y = \dfrac{100}{1+2^{-x}}$
Take the reciprocal: $\displaystyle \dfrac{1}{y} = \dfrac{1+2^{-x}}{100}$
Multiply both sides by 100: $\displaystyle \dfrac{100}{y} = 1+2^{-x}$
Subtract 1 from both sides: $\displaystyle 2^{-x} = \dfrac{100}{y} - 1 = \dfrac{100-y}{y}$
It should be relatively easy to solve for x using logarithms. You can pick whichever base you like although I prefer base e (the natural log)
Using logarithms to get this is the best method of solving exponential equations like this, perhaps you should revise them when you next get a chance.
Since the book wants to use base 2 lets go with that: $\displaystyle \log_2(2^{-x}) = \log_2 \left(\dfrac{100-y}{y}\right)$
By our log laws we know that $\displaystyle \log_a(a) = 1 \text{ and } \log_c(a^k) = k \log_c(a)$ so we can rewrite the LHS as $\displaystyle \log_2(2^{-x}) = -x \log_2(2) = -x$. Do you understand how I used the laws to get to -x ?
In the equation as whole we have, after dividing by -1: $\displaystyle x = -\log_2 \left(\dfrac{100-y}{y}\right)$
Now that we have x in terms of y rewrite x as $\displaystyle f^{-1}(x)$ and y as $\displaystyle x$:
$\displaystyle f^{-1}(x) = -\log_2 \left(\dfrac{100-x}{x}\right)$.
This is the same as your book's answer, just written in a different way. If you want the book's answer use the log power laws on that -1 outside the logarithm and the effect of raising a fraction to power -1
If you have given:
$\displaystyle y=\frac{100}{1+2^{-x}}$
and you've to find $\displaystyle f^{-1}(x)$ then change $\displaystyle x$ en $\displaystyle y$:
$\displaystyle x=\frac{100}{1+2^{-y}} \Leftrightarrow 2^{-y}=\frac{100}{x}-1$
Now take the log with base 2 from each side:
$\displaystyle \log_2\left(2^{-y}\right)=\log_2\left(\frac{100-x}{x}\right)$
$\displaystyle \Leftrightarrow -y=\log_2\left(\frac{100-x}{x}\right)$
$\displaystyle \Leftrightarrow y= -\log_2\left(\frac{100-x}{x}\right)$
$\displaystyle \Leftrightarrow y=\log_2\left(\frac{x}{100-x}\right)$