1. ## Inverse 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??

2. ## 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:

$y = \dfrac{100}{1+2^{-x}}$

Take the reciprocal: $\dfrac{1}{y} = \dfrac{1+2^{-x}}{100}$

Multiply both sides by 100: $\dfrac{100}{y} = 1+2^{-x}$

Subtract 1 from both sides: $2^{-x} = \dfrac{100}{y} - 1 = \dfrac{100-y}{y}$

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)

3. ## Re: 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?

4. ## Re: I need help with an inverse logarithm problem!!

How far can you get from where I left in my previous post?

5. ## Re: I need help with an inverse logarithm problem!!

Not very far. I want to get rid of the 2^-x somehow.

6. ## Re: I need help with an inverse logarithm problem!!

Originally Posted by ltlbit94
Not very far. I want to get rid of the 2^-x somehow.
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: $\log_2(2^{-x}) = \log_2 \left(\dfrac{100-y}{y}\right)$

By our log laws we know that $\log_a(a) = 1 \text{ and } \log_c(a^k) = k \log_c(a)$ so we can rewrite the LHS as $\log_2(2^{-x}) = -x \log_2(2) = -x$. Do you understand how I used the laws to get to -x ?

In the equation as whole we have, after dividing by -1: $x = -\log_2 \left(\dfrac{100-y}{y}\right)$

Now that we have x in terms of y rewrite x as $f^{-1}(x)$ and y as $x$:

$f^{-1}(x) = -\log_2 \left(\dfrac{100-x}{x}\right)$.

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

7. ## Re: I need help with an inverse logarithm problem!!

If you have given:
$y=\frac{100}{1+2^{-x}}$
and you've to find $f^{-1}(x)$ then change $x$ en $y$:
$x=\frac{100}{1+2^{-y}} \Leftrightarrow 2^{-y}=\frac{100}{x}-1$
Now take the log with base 2 from each side:
$\log_2\left(2^{-y}\right)=\log_2\left(\frac{100-x}{x}\right)$
$\Leftrightarrow -y=\log_2\left(\frac{100-x}{x}\right)$
$\Leftrightarrow y= -\log_2\left(\frac{100-x}{x}\right)$
$\Leftrightarrow y=\log_2\left(\frac{x}{100-x}\right)$

8. ## Re: Inverse logarithm problem!

Originally Posted by ltlbit94
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??
From the content of a pm from the OP, it seems this question is part of an assignment that possibly counts towards the student's grade. See rule 6. Thread closed.