# Proof/Show using index laws

1. Mar 1, 2013

1. The problem statement, all variables and given/known data
Show that the following can be expressed as a quadratic equation

$$2^{x}-1 = \frac{2}{2^{x}}$$

2. Relevant equations
Show that the quadratic has only one real solution at $x = 1$

3. The attempt at a solution
Well, my biggest problem is the main part of the task, converting the equation into a quadratic. I cannot find a way to get the same base for all parts of the equation so I can equate them. I've tried doing things like:
-Taking the denominator to the top on the RHS and making $1 = 2^{0}$
- Squaring both sides of the equation, although I can get a quadratic out of this method, technically I haven't proved anything, just made the equation look a little fancier.
- taking the -1 to the other side and adding it in the form of $\frac{2^{x}}{2^{x}}$

And others too, but those were the ones which actually led me somewhere (but that somewhere was a dead end).

Can anyone give me any hints on how to approach a question like this?

ps. I've also tried not to assume x = 1 while doing my conversion to a quadratic.

Last edited by a moderator: Mar 1, 2013
2. Mar 1, 2013

### rock.freak667

The 2^x in the denominator doesn't look too helpful in that position, so make it look nicer and multiply throughout by 2^x

3. Mar 1, 2013

### HallsofIvy

Staff Emeritus
Much the same thing: replace each "$2^x$" with "y" and then multiply both sides by y.

4. Mar 2, 2013

I tried the multiply y to both sides, but I end up getting something like this:

$\frac{2^{2x}-2^{x}}{2^{x}}=\frac{2^{x+1}}{2^{2x}}$

Which is where I get stuck again. Although I get a quadratic when I sub in the y, when I revert it back to $2^{x}$, the function becomes a plain old exponential function.

If I split the LHS at the "-" I'll still get the pesky -1 and ending up cancelling out one of the x, which put me at a dead end

How do I proceed?

5. Mar 2, 2013

### MrWarlock616

Hey you did the multiplication wrong! $2^{x+1}$??

It should be: $2^{2x}-2^x=2$. Now a simple substitution of 2x=y will give you a quadratic equation..

6. Mar 2, 2013

I multiplied both sides by
$\frac{2^{x}}{2^{x}}$ that's how I got that result

Yeah but is that really a quadratic equation? or an exponential function in disguise. Because if I have my quadratic in terms of y, as soon as I sub in $2^{x}$ it turns into an exponential function. I checked this by graphing it, and it does not pass through x = 1

I graphed:
$2^{2x}-2^{x}+2 = y$

7. Mar 2, 2013

### MrWarlock616

No! You don't have to graph that.
Solve: $y^2-y-2=0$, where y=2x..

Then you get 2 values for $2^x$, using which you can prove the next part of the question.

8. Mar 2, 2013