#### 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:

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#### Hurkyl

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

#### cscott

Hurkyl said:
Maybe you haven't done anything wrong? What would that imply?
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.

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#### 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?

#### cscott

wisredz said:
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?
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.

#### Nylex

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

#### cscott

Nylex said:
What he's saying is that there is no value for x that satisfies that equation.
Why would this be on my grade 11 math exam then?

#### Nylex

cscott said:
Why would this be on my grade 11 math exam then?
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.

#### saltydog

Homework Helper
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.

#### Hurkyl

Staff Emeritus
Gold Member
Why would this be on my grade 11 math exam then?
Presumably to test if you can identify when equations have no solutions. :tongue2:

#### cscott

Hurkyl said:
Presumably to test if you can identify when equations have no solutions. :tongue2:
Crazy! Anyway, thanks for all your help.

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