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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lynt.br on October 05, 2009, 05:17:52 pm

Title: Quick question
Post by: lynt.br on October 05, 2009, 05:17:52 pm: Could someone please explain why $\sqrt[a]{x^b} = (\sqrt[a]{x})^b$

Thanks in advance.
Title: Re: Quick question
Post by: shinny on October 05, 2009, 05:26:07 pm: $\sqrt[a]{x^b}=(x^b)^{\frac{1}{a}}=x^{\frac{b}{a}}=(x^{\frac{1}{a}})^{b}=(\sqrt[a]{x})^{b}$

I'm sure others could show it using a much prettier proof, but that's basically showing it through the index law of multiplying powers when you've got a power to another power.
Title: Re: Quick question
Post by: Flaming_Arrow on October 05, 2009, 05:26:39 pm: $\sqrt[a]{x^b} = x^\frac{b}{a}$

$\sqrt[a]{x} = x^\frac{1}{a}$

$\therefore\sqrt[a]{x^b}= (\sqrt[a]{x})^b$
Title: Re: Quick question
Post by: /0 on October 05, 2009, 05:32:11 pm: For $x \geq 0$
Title: Re: Quick question
Post by: TrueTears on October 05, 2009, 05:35:49 pm: Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
Serious? I thought you could have $(-2)^{\frac{6}{3}}$
Title: Re: Quick question
Post by: lynt.br on October 05, 2009, 05:47:16 pm: Cheers guys, makes sense now.
Title: Re: Quick question
Post by: kamil9876 on October 05, 2009, 06:10:37 pm: notice that for natural numbers:

$(c_1c_2c_3...c_b)^a=c_1^ac_2^ac_3^a....c_b^a$ (there are $b$ numbers as seen from the subscript). (ask for proof of this if not convinced, I will leave that as a seperate matter)

If $^a\sqrt{x}=c_1=c_2=....=c_b$

Then the first equation gives:

$(^a\sqrt{x} ^a\sqrt{x} ^a\sqrt{x}....)^a=(^a\sqrt{x})^a(^a\sqrt{x})^a(^a\sqrt{x})^a.....$ ( $b$ terms)

$((^a\sqrt{x})^b)^a=xxxxxxxx...$

$((^a\sqrt{x})^b)^a=x^b$

$a^{th}$ rooting both sides:

$(^a\sqrt{x})^b=^a\sqrt{x^b}$

Fractional exponents are the much easier way, and in fact you should use them and forget about ath root notation, since the notation of fractional exponents has so much theory loaded into it that it suggests all these non-trivial properties as trivial. However the above argument provides the logical foundations for using fractional exponents and gives motivation for developing theory of them.
Title: Re: Quick question
Post by: /0 on October 06, 2009, 09:13:40 pm: Quote from: TrueTears on October 05, 2009, 05:35:49 pm
Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
Serious? I thought you could have $(-2)^{\frac{6}{3}}$

While true for certain values of $a$ , $b$ . In general you can't conclude that

$\sqrt[a]{x^b} = (\sqrt[a]{x})^b$ for $x < 0$

e.g.

$\sqrt[2]{(-2)^2}\neq(\sqrt[2]{-2})^2$
Title: Re: Quick question
Post by: TrueTears on October 06, 2009, 09:15:30 pm: Quote from: /0 on October 06, 2009, 09:13:40 pm
Quote from: TrueTears on October 05, 2009, 05:35:49 pm
Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
Serious? I thought you could have $(-2)^{\frac{6}{3}}$

While true for certain values of $a$ , $b$ . In general you can't conclude that

$\sqrt[a]{x^b} = (\sqrt[a]{x})^b$ for $x < 0$

e.g.

$\sqrt[2]{(-2)^2}\neq(\sqrt[2]{-2})^2$
But if true for certain values then you can not say for all $x \ge 0$

EDIT: "say"
Title: Re: Quick question
Post by: /0 on October 06, 2009, 09:16:18 pm: Quote from: TrueTears on October 06, 2009, 09:15:30 pm
Quote from: /0 on October 06, 2009, 09:13:40 pm
Quote from: TrueTears on October 05, 2009, 05:35:49 pm
Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
Serious? I thought you could have $(-2)^{\frac{6}{3}}$

While true for certain values of $a$ , $b$ . In general you can't conclude that

$\sqrt[a]{x^b} = (\sqrt[a]{x})^b$ for $x < 0$

e.g.

$\sqrt[2]{(-2)^2}\neq(\sqrt[2]{-2})^2$
But if true for certain values then you can not for all $x \ge 0$

What do you mean?
Title: Re: Quick question
Post by: TrueTears on October 06, 2009, 09:19:53 pm: Quote from: Flaming_Arrow on October 05, 2009, 05:26:39 pm
$\sqrt[a]{x^b} = x^\frac{b}{a}$

$\sqrt[a]{x} = x^\frac{1}{a}$

$\therefore\sqrt[a]{x^b}= (\sqrt[a]{x})^b$
Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
But if for certain values of x satisfies the expression, then you can't say x must be $\ge 0$
Title: Re: Quick question
Post by: /0 on October 06, 2009, 09:25:14 pm: I didn't say that you must have $x \geq 0$ . It is simply a subset of the values $x$ can take.
However, if you want to make $x$ negative then you must simultaneously introduce restrictions on $a$ and $b$ which can make things very tricky.

You cannot simply say

$\sqrt[a]{x^b} = (\sqrt[a]{x})^b$ for all x if you get my drift.
Title: Re: Quick question
Post by: TrueTears on October 06, 2009, 09:30:55 pm: Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
implies you mean for all $x \ge 0$

or why would you even mention a subset of the domain?

It's like saying you got $y = \sqrt{x}$ with no restriction yet I'm gonna put $x \ge 5$ after it since it's a subset.

Soz I'm just not getting the point if you know what I mean.
Title: Re: Quick question
Post by: /0 on October 06, 2009, 09:45:56 pm: Quote from: TrueTears on October 06, 2009, 09:30:55 pm
Quote from: /0 on October 05, 2009, 05:32:11 pm
For $x \geq 0$
implies you mean for all $x \ge 0$

Gee, I think you're right about that.

It is defined for all $x \ge 0$ right, so there's nothing wrong about that?

It is all we will need to worry in VCE, and if you want to write up the restrictions for $a$ and $b$ , be my guest.
Title: Re: Quick question
Post by: TrueTears on October 06, 2009, 09:48:41 pm: Yes it is defined but I don't see the point of writing it, like I said $y = \sqrt{x}$ is defined for $x \ge 5$ right? But why would you even put a subset after the expression.
Title: Re: Quick question
Post by: /0 on October 06, 2009, 09:52:16 pm: Just so you people know that it isn't always defined for $x < 0$ ... I'm not saying Flaming Arrow needed to or anything like that, I was just adding information about the values of $x$ for which the equation is guaranteed to work
Title: Re: Quick question
Post by: TrueTears on October 06, 2009, 09:54:09 pm: Oh, I thought you meant it as the implied domain.

ie, $y = \sqrt{x}$ , $x \ge 0$