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Quadratic Exponential question

  • Thread starter cscott
  • Start date
783
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This was on a test but I couldn't quite solve for x:

[tex]5^2^x + 4(5)^x = -3[/tex]

let [itex]5^x = y[/itex]

[tex]y^2 + 4y + 3 = 0[/tex]
[tex](y + 1)(y + 3) = 0[/tex]

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

Hurkyl

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Maybe you haven't done anything wrong? What would that imply?
 
783
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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. :biggrin:

Overall, I don't know what this means.
 
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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?
 
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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.
 
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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.
 
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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? :mad:
 
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cscott said:
Why would this be on my grade 11 math exam then? :mad:
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

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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

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Why would this be on my grade 11 math exam then?
Presumably to test if you can identify when equations have no solutions. :tongue2:
 
783
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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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