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??

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- Aug 7th 2011, 10:24 AMltlbit94Inverse logarithm problem!
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??

- Aug 7th 2011, 10:37 AMe^(i*pi)Re: I need help with an inverse logarithm problem!!
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) - Aug 7th 2011, 10:49 AMltlbit94Re: I need help with an inverse logarithm problem!!
But the back of my book gives me the answer f^-1(x)=log base 2(x/100-x). How do I get to that answer?

- Aug 7th 2011, 11:00 AMe^(i*pi)Re: I need help with an inverse logarithm problem!!
How far can you get from where I left in my previous post?

- Aug 7th 2011, 11:07 AMltlbit94Re: I need help with an inverse logarithm problem!!
Not very far. I want to get rid of the 2^-x somehow.

- Aug 7th 2011, 11:16 AMe^(i*pi)Re: I need help with an inverse logarithm problem!!
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 - Aug 7th 2011, 11:20 AMSironRe: I need help with an inverse logarithm problem!!
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:

$\displaystyle \log_2\left(2^{-y}\right)=\log_2\left(\frac{100-x}{x}\right)$

$\displaystyle \Leftrightarrow -y=\log_2\left(\frac{100-x}{x}\right)$

$\displaystyle \Leftrightarrow y= -\log_2\left(\frac{100-x}{x}\right)$

$\displaystyle \Leftrightarrow y=\log_2\left(\frac{x}{100-x}\right)$ - Aug 7th 2011, 11:12 PMmr fantasticRe: Inverse logarithm problem!