# Homework Help: Quadratic Exponential question

1. Jun 12, 2005

### cscott

This was on a test but I couldn't quite solve for x:

$$5^2^x + 4(5)^x = -3$$

let $5^x = y$

$$y^2 + 4y + 3 = 0$$
$$(y + 1)(y + 3) = 0$$

So I end up with $5^x = -1$ or $5^x = -3$, but I don't think that makes sense... what am I doing wrong? It must be something dumb I'm doing :shy:

Last edited: Jun 12, 2005
2. Jun 12, 2005

### Hurkyl

Staff Emeritus
Maybe you haven't done anything wrong? What would that imply?

3. Jun 12, 2005

### cscott

Ehh I was kind of worried about that... I guess it could imply I suck at logs? :tongue:

I would continue by taking the log of both sides, but as far as I know you cannot take the log of a negative number... my calculator agrees with me.

Overall, I don't know what this means.

Last edited: Jun 12, 2005
4. Jun 12, 2005

### wisredz

It meons there's no real number x that is the root of this function... Think of it this way, which power of 5(a positive number) would give you -1 or -3?

5. Jun 12, 2005

### cscott

How would I work it out to get an answer? All we've been taught on complex numbers (if that's what you're implying) is that i is the square root of negative one.

6. Jun 12, 2005

### Nylex

What he's saying is that there is no value for x that satisfies that equation.

7. Jun 12, 2005

### cscott

Why would this be on my grade 11 math exam then?

8. Jun 12, 2005

### Nylex

I don't know, but surely you can see that there's no power x that you can raise 5 to to get a negative number, not even a negative one. Your working at the top was correct.

9. Jun 12, 2005

### saltydog

Well it's an interesting problem for me. There are no solutions with "real numbers" but there are two complex-number solutions which perhaps it's best to not worry about now unless you want to know how to find them.

10. Jun 12, 2005

### Hurkyl

Staff Emeritus
Presumably to test if you can identify when equations have no solutions. :tongue2:

11. Jun 12, 2005

### cscott

Crazy! Anyway, thanks for all your help.