# Inverse of 3^(-x) + 9^(-x)

1. Jun 21, 2012

### ishant

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

y = 3^-x + 9^-x

ln(y) = ln(3^-x + 9^-x)

don't know where to go from here.

2. Jun 21, 2012

### Fightfish

Basically you want to make x the subject.

Hint 1: What can you rewrite 9^(-x) as?
Hint 2: Solve for 3^(-x) by formulating the problem as a quadratic equation

3. Jun 21, 2012

### SammyS

Staff Emeritus
Logarithms won't help initially.

Follow Fightfish's advice directly, or first let u = 3-x, so u2 = (3-x)2 =   ?   ... and then look at this as a quadratic equation.

4. Jun 22, 2012

### HallsofIvy

Staff Emeritus
And because it reduces to a quadratic, this function does NOT have a true "inverse". You can reduce the domain to two intervals so the restricted functions have inverses.

5. Jun 22, 2012

### InfinityZero

But since 3-x is always positive we can eliminate one of the solutions to the quadratic and be left with a unique inverse, right? This is of course only if the domain only includes real numbers.

6. Jun 22, 2012

### SammyS

Staff Emeritus
Yes, that's correct, at least for this function.

The fact that this function, f(x) = 3-x + 9-x, does have a true inverse can also be seen by noticing that f(x) is strictly decreasing.

7. Jun 23, 2012

### SammyS

Staff Emeritus
While we wait for OP (ishant) to return, i''l take the opportunity to amplify my above reply to the above quotes from HallsofIvy and InfinityZero.

'Halls', of course, is correct in general. A quadratic function does not have a true inverse, unless the domain of the quadratic function is restricted appropriately.

InfinityZero was correct about the function, f(x) = 3-x + 9-x, having a unique inverse, that is to say, an inverse which is truly a function. However, the reason 'IZ' gives, could stand to be elaborated upon. The reason that we can eliminate one of the two solutions to the quadratic equation which results when solving for 3-x, is that 3-x is a positive quantity and one of the solutions is positive while the other is strictly negative.

Let's look at a slightly different function for an example of what 'Halls' cautioned about.
Suppose we want to solve y = 9-x - 3-x for x.

Let F(x) = 9-x - 3-x.

F is not 1 to 1. It has a minimum of -1/4 at x = log3(2) .

Now to solve y = 9-x - 3-x for x:

Let t = 3-x. Substituting that into the equation for y gives:

t2 - t = y .

Add 1/4 to both sides to complete the square giving:

(t - 1/2)2 = y + 1/4

Solving for t gives us:

$\displaystyle t=\frac{1}{2}\pm\sqrt{y+\frac{1}{4}}$

Since x = -log3(t), our solution is:

$\displaystyle x=-\log_{\,3}\left(\frac{1}{2}\pm\sqrt{y+\frac{1}{4}} \right)$

If we restrict the domain of F to x ≤ log3(2), then F-1(x) is the result with the minus sign.

$\displaystyle F^{-1}(y)=-\log_{\,3}\left(\frac{1}{2}-\sqrt{y+\frac{1}{4}} \right)$

This is valid for any y such that y ≥ -1/4 .

On the other hand, if we restrict the domain of F to x ≥ log3(2), then F-1(x) is the result with the plus sign.

$\displaystyle F^{-1}(y)=-\log_{\,3}\left(\frac{1}{2}+\sqrt{y+\frac{1}{4}} \right)$

This is valid for any y such that -1/4 ≤ y < 0 .​
Here is a graph of F(x) = 9-x - 3-x as given by WolframAlpha.

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8. Jun 23, 2012

### vela

Staff Emeritus
I deleted the subthread about the approximate solutions as it will likely only serve to confuse the OP.