How does -2^(5/2) -2^(5/2) = -2^(7/2)?

[Rosebud]

Homework Statement

How does -2^(5/2) -2^(5/2) = -2^(7/2)?

Homework Equations

I've integrated a problem down to this and I know that the answer is -2^(7/2). Unfortunately, I've forgotten the algebraic steps required to get it into that form.

The Attempt at a Solution

I'm totally lost.

 

[NihilTico]

[itex](-2)^{5/2}+(-2)^{5/2}=2(-2)^{5/2}[/itex]

It might help to think about it like this. We write [itex]x=y^{a/b}[/itex] to mean roughly that [itex]x[/itex] is the number equal to [itex]y[/itex] multiplied together with itself "[itex]a/b[/itex] times". In this sense, how might you think we should write [itex]2(-2)^{5/2}[/itex]?

 

[Rosebud]

[itex](-2)^{5/2}+(-2)^{5/2}=2(-2)^{5/2}[/itex]

It might help to think about it like this. We write [itex]x=y^{a/b}[/itex] to mean roughly that [itex]x[/itex] is the number equal to [itex]y[/itex] multiplied together with itself "[itex]a/b[/itex] times". In this sense, how might you think we should write [itex]2(-2)^{5/2}[/itex]?

Is this logic correct, (-)2^(1+5/2) = (-)2^(2/2)+(5/2) = -2^(7/2)? Can I do that with the negative sign? It doesn't seem like I can.

 

[NihilTico]

Is this logic correct, (-)2^(1+5/2) = (-)2^(2/2)+(5/2) = -2^(7/2)? Can I do that with the negative sign? It doesn't seem like I can.

If the negative sign is outside the power, you can pull it out all together until you're done simplifying. If the negative sign is being raised to a power as well, you must bring it along.

So, if you have the negative signs all out front, and not being raised to a power, then yes, because [itex]-x=(-1)\cdot{x}[/itex]. So what you've done is precisely correct in that case, except for the notation. Writing (-) in any situation doesn't make any sense. We use [itex]-n[/itex] (for some number [itex]n[/itex]) to basically mean less than [itex]0[/itex] in the standard ordering of the real numbers (i.e., [itex]3\le\pi[/itex] etc.). In other words, I would write your answer as:

-2^(5/2)-2^(5/2)=-(2^(5/2)+2^(5/2))=-(2^(1+5/2))=-(2^((2/2)+(5/2)))=-2^(7/2)

Or in TeX

[itex]-2^{(5/2)}-2^{(5/2)}=-(2^{(5/2)}+2^{(5/2)})=-(2^{(1+5/2)})=-(2^{(2/2)+(5/2)})=-2^{(7/2)}[/itex]

 

[Mentallic]

How does -2^(5/2) -2^(5/2) = -2^(7/2)?

[itex](-2)^{5/2}+(-2)^{5/2}=2(-2)^{5/2}[/itex]

NihilTico, what you wrote is different to what the OP has. [itex](-2)^n=(-1)^n2^n[/itex] while [itex]-2^n=-(2^n)[/itex]. The first is a positive number when n is an even integer, and negative when n is odd. It is also a complex number when n is neither of those. The second expression however is always a negative number.

Is this logic correct, (-)2^(1+5/2) = (-)2^(2/2)+(5/2) = -2^(7/2)? Can I do that with the negative sign? It doesn't seem like I can.

Yes, that is correct, but you don't need to surround the negative sign in brackets. I'd write it like this:

[tex]-2^{5/2}-2^{5/2}[/tex]
[tex]=2(-2^{5/2})[/tex]
[tex]=-2^12^{5/2}[/tex]
[tex]=-2^{1+5/2}[/tex]
[tex]-2^{2/2+5/2}[/tex]
[tex]=-2^{7/2}[/tex]

But of course when you get more accustomed to the rules, you can skip a lot of these steps.

 

[NihilTico]

NihilTico, what you wrote is different to what the OP has.

Well, yes, I figured that when Rosebud replied with the same notation ;)

 

[Rosebud]

Thank you both for your time and effort. I can finally go to sleep now, lol.