Exponents and Radicals confusing transition

[potatojanitor]

Hi all, I'm prepping for the SAT and I'm using Khan Academy SAT Prep to hone my math. I'm doing "Exponents and rational exponents" practice and I've run into a snag. I tried to answer but in the end I decided to guess, so I picked B. What I'm stuck with is the explanation. Specifically, the part where it simplified the expression, which I circled in the image. I don't understand how you can make the c in the denominator disappear like that. I know that when a^b ÷ a^c = a^(b-c), but why is it gone? Is it because when c⁰ = 1? And 1 • 4 = 4? Even then, it seems flawed to do it like that. Is it because (8/32) • (c²⁰/c⁴) = (1/4) • (c¹⁶/c⁰) = (1/4) • (c¹⁶/1)= (1/4) • c¹⁶ = c¹⁶/4? Or is it (8/32) • (c²⁰/c⁴) = (1/4) • c^{(20 - 4)} = (1/4) • c¹⁶ = c¹⁶/4? I use the parentheses to separate the thingys. Is this correct?

 

[lev888]

`2^-3 = 1/2^3`
Agree? That's how it "moves" between numerator and denominator.

 

[potatojanitor]

So after tackling with this for more than an hour, I figured out how to use LaTex and I think my explanation is right. For some reason, I can't seem to delete this post, so apologies to the people who just spent some time trying to read it! However, I'm still not sure if this is correct so if you could let me know that'd be great!
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}} = \sqrt{\frac{8}{32}\cdot \frac{c^{20}}{c^{4}}} = \sqrt{\frac{1}{4}\cdot c^{16}} = \sqrt{\frac{c^{16}}{4}} = \frac{\sqrt{c^{16}}}{\sqrt{4}} = \frac{\left ( c^{16} \right )^{\frac{1}{2}}}{2} = \frac{c^{8}}{2} = \frac{1}{2} c^{8}[/MATH]

 

[Deleted member 4993]

So I figured out how to use LaTex and I think my math is right and I understand it. For some reason, I can't seem to delete this post, so apologies to the people who just spent some time trying to read it! However, I'm still not sure if this is correct so if you could let me know that'd be great!
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}} = \sqrt{\frac{8}{32}\cdot \frac{c^{20}}{c^{4}}} = \sqrt{\frac{1}{4}\cdot c^{16}} = \sqrt{\frac{c^{16}}{4}} = \frac{\sqrt{c^{16}}}{\sqrt{4}} = \frac{\left ( c^{16} \right )^{\frac{1}{2}}}{2} = \frac{c^{8}}{2} = \frac{1}{2} c^{8}[/MATH]

Looks gpod!!

 

[potatojanitor]

2−3=1232-3=123
Agree? That's how it "moves" between numerator and denominator.

Thanks my dude, but I am yet to reach your level of intellect to understand your response. Thanks tho!

 

[potatojanitor]

Looks gpod!!

Thanks!

 

[lev888]

You wrote:

I know that when `a^b ÷ a^c = a^(b-c)`, but why is it gone?

So it seems you don't understand how `a^c` moved from denominator to numerator, right? I tried to show how it happens. By definition, negative exponent is 1 divided by positive exponent:
`a^-c = 1/a^c`
And the other way around:
`1/a^c = a^-c`
So, `a^b/a^c = a^b * a^-c = a^(b-c)`

 

[potatojanitor]

You wrote:

So it seems you don't understand how `a^c` moved from denominator to numerator, right? I tried to show how it happens. By definition, negative exponent is 1 divided by positive exponent:
`a^-c = 1/a^c`
And the other way around:
`1/a^c = a^-c`
So, `a^b/a^c = a^b * a^-c = a^(b-c)`

Exactly. At first glance, I didn't understand how c, from the numerator, could directly subtract c, from the denominator. This is the way I thought it was done initially:
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}} = \sqrt{\frac{8\cdot c^{20}}{32\cdot c^{4}}} = \sqrt{\frac{1\cdot c^{20}}{4\cdot c^{4}}}[/MATH]And then, somehow, the c's magically deduct the exponents like we're doing subtraction.
[MATH]\sqrt{\frac{1\cdot c^{20}}{4\cdot c^{4}}} = \sqrt{\frac{1\cdot c^{16}}{4\cdot c^{0}}} = \sqrt{\frac{1\cdot c^{16}}{4\cdot 1}} = \sqrt{\frac{c^{16}}{4}}[/MATH]That was probably incorrect. I understand what you mean because applying [MATH]\frac{a^{b}}{a^{c}} = a^{b} \cdot a^{-c} = a^{b} \cdot \frac{1}{a^{c}}[/MATH], it would be this:
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}}= \sqrt{\frac{8}{32}\cdot\frac{c^{20}}{c^{4}}}[/MATH]And then applying the rules you mentioned:
[MATH]\sqrt{\frac{1}{4}\cdot\frac{c^{20}}{c^{4}}} = \sqrt{\frac{1}{4}\cdot\left ( c^{20} \cdot \frac{1}{c^{4}} \right )} = \sqrt{\frac{1}{4}\cdot \left ( c^{20} \cdot c^{-4} \right )} = \sqrt{\frac{1}{4}\cdot c^{16}} = \sqrt{\frac{c^{16}}{4}}[/MATH]Thanks for the help!

 

[lev888]

Exactly. At first glance, I didn't understand how c, from the numerator, could directly subtract c, from the denominator. This is the way I thought it was done initially:
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}} = \sqrt{\frac{8\cdot c^{20}}{32\cdot c^{4}}} = \sqrt{\frac{1\cdot c^{20}}{4\cdot c^{4}}}[/MATH]And then, somehow, the c's magically deduct the exponents like we're doing subtraction.
[MATH]\sqrt{\frac{1\cdot c^{20}}{4\cdot c^{4}}} = \sqrt{\frac{1\cdot c^{16}}{4\cdot c^{0}}} = \sqrt{\frac{1\cdot c^{16}}{4\cdot 1}} = \sqrt{\frac{c^{16}}{4}}[/MATH]That was probably incorrect. I understand what you mean because applying [MATH]\frac{a^{b}}{a^{c}} = a^{b} \cdot a^{-c} = a^{b} \cdot \frac{1}{a^{c}}[/MATH], it would be this:
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}}= \sqrt{\frac{8}{32}\cdot\frac{c^{20}}{c^{4}}}[/MATH]And then applying the rules you mentioned:
[MATH]\sqrt{\frac{1}{4}\cdot\frac{c^{20}}{c^{4}}} = \sqrt{\frac{1}{4}\cdot\left ( c^{20} \cdot \frac{1}{c^{4}} \right )} = \sqrt{\frac{1}{4}\cdot \left ( c^{20} \cdot c^{-4} \right )} = \sqrt{\frac{1}{4}\cdot c^{16}} = \sqrt{\frac{c^{16}}{4}}[/MATH]Thanks for the help!

The magical subtraction of exponents IS correct. You can do it without moving things up or down first.

 

[jonah2.0]

Beer soaked ramblings follow.
[/QUOTE]

So after tackling with this for more than an hour, I figured out how to use LaTex and I think my explanation is right. For some reason, I can't seem to delete this post, so apologies to the people who just spent some time trying to read it! However, I'm still not sure if this is correct so if you could let me know that'd be great!
[MATH]\sqrt{\frac{8c^{20}}{32c^{4}}} = \sqrt{\frac{8}{32}\cdot \frac{c^{20}}{c^{4}}} = \sqrt{\frac{1}{4}\cdot c^{16}} = \sqrt{\frac{c^{16}}{4}} = \frac{\sqrt{c^{16}}}{\sqrt{4}} = \frac{\left ( c^{16} \right )^{\frac{1}{2}}}{2} = \frac{c^{8}}{2} = \frac{1}{2} c^{8}[/MATH]

Man, that's fast.
Latex fast.

 

[JeffM]

Assume a is a real number > 0.

[MATH]a^p * a^q = a^{(p + q)}.[/MATH] Right?

[MATH]a^{-r} = \dfrac{1}{a^r}.[/MATH] Right?

Now lets do a little algebra.

[MATH]x = a^s * a^{-t}.[/MATH]
[MATH]\text {Let } u = - t \implies x = a^s * a^u = a^{(s + u)} = a^{(s-t)}.[/MATH]
[MATH]x = a^s * a^{(-t)} = a^s * \dfrac{1}{a^t} = \dfrac{a^s}{1} * \dfrac{1}{a^t} = \dfrac{a^s}{a^t}.[/MATH]
[MATH]x = \dfrac{a^s}{a^t} \text { and } x = a^{(s-t)} \implies a^{(s-t)} = \dfrac{a^s}{a^t}.[/MATH]
Let’s call that theorem A.

Assume b and c are real numbers > 0, and d is an integer > 1.

[MATH]y = \sqrt[d]{b} * \sqrt[d]{c} \implies y^d = (\sqrt[d]{b} * \sqrt[d]{c})^d = bc \implies y = \sqrt[d]{bc}.[/MATH]
[MATH]y = \sqrt[d]{b} * \sqrt [d]{c} \text { and } y = \sqrt[d]{bc} \implies \sqrt[d]{b} * \sqrt[d]{c} = \sqrt[d]{bc}[/MATH]
Let’s call that Theorem B.

[MATH]w = \dfrac{\sqrt[d]{c}}{\sqrt[d]{c}} \implies w^d = \left ( \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} \right )^d = \dfrac{b}{c} \implies w = \sqrt[d]{\dfrac{b}{c}}.[/MATH]
[MATH]w = \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} \text { and } w = \sqrt[d]{\dfrac{b}{c}} \implies \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} = \sqrt[d]{\dfrac{b}{c}}.[/MATH]
Call that Theorem C.

Your problem requires Theorems A, B, and C.

[MATH]\dfrac{\sqrt{8c^{20}}}{\sqrt{32c^4}} = \sqrt{\dfrac{8c^{20}}{32c^4}} = \sqrt{\dfrac{1}{4} * c^{(20-4)}} = \sqrt{\dfrac{1}{4}} * \sqrt{c^{16}} = \dfrac{1}{2} * c^8.[/MATH]

 

[potatojanitor]

Assume a is a real number > 0.

[MATH]a^p * a^q = a^{(p + q)}.[/MATH] Right?

[MATH]a^{-r} = \dfrac{1}{a^r}.[/MATH] Right?

Now lets do a little algebra.

[MATH]x = a^s * a^{-t}.[/MATH]
[MATH]\text {Let } u = - t \implies x = a^s * a^u = a^{(s + u)} = a^{(s-t)}.[/MATH]
[MATH]x = a^s * a^{(-t)} = a^s * \dfrac{1}{a^t} = \dfrac{a^s}{1} * \dfrac{1}{a^t} = \dfrac{a^s}{a^t}.[/MATH]
[MATH]x = \dfrac{a^s}{a^t} \text { and } x = a^{(s-t)} \implies a^{(s-t)} = \dfrac{a^s}{a^t}.[/MATH]
Let’s call that theorem A.

Assume b and c are real numbers > 0, and d is an integer > 1.

[MATH]y = \sqrt[d]{b} * \sqrt[d]{c} \implies y^d = (\sqrt[d]{b} * \sqrt[d]{c})^d = bc \implies y = \sqrt[d]{bc}.[/MATH]
[MATH]y = \sqrt[d]{b} * \sqrt [d]{c} \text { and } y = \sqrt[d]{bc} \implies \sqrt[d]{b} * \sqrt[d]{c} = \sqrt[d]{bc}[/MATH]
Let’s call that Theorem B.

[MATH]w = \dfrac{\sqrt[d]{c}}{\sqrt[d]{c}} \implies w^d = \left ( \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} \right )^d = \dfrac{b}{c} \implies w = \sqrt[d]{\dfrac{b}{c}}.[/MATH]
[MATH]w = \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} \text { and } w = \sqrt[d]{\dfrac{b}{c}} \implies \dfrac{\sqrt[d]{b}}{\sqrt[d]{c}} = \sqrt[d]{\dfrac{b}{c}}.[/MATH]
Call that Theorem C.

Your problem requires Theorems A, B, and C.

[MATH]\dfrac{\sqrt{8c^{20}}}{\sqrt{32c^4}} = \sqrt{\dfrac{8c^{20}}{32c^4}} = \sqrt{\dfrac{1}{4} * c^{(20-4)}} = \sqrt{\dfrac{1}{4}} * \sqrt{c^{16}} = \dfrac{1}{2} * c^8.[/MATH]

Why thank you, JeffM. That was very insightful, especially theorem B & C because I've never thought about it that way. I didn't know I could take away the radical by raising the other factors to a power.

 

[JeffM]

Why thank you, JeffM. That was very insightful, especially theorem B & C because I've never thought about it that way. I didn't know I could take away the radical by raising the other factors to a power.

You are welcome. For some reason, teachers at the beginning level do no explain the power of generalization. I am not sayibg that most teachers at the beginning level do not understand it. Rather, I suspect that they MISTAKENLY view the power of generalization to too obvious to mention.

I read some wise mathematician who said every lesson should teach the power of generalization and abstraction. I am not sure it is wise about everything, spouses for example, but it is true about math.

 

[Deleted member 4993]

Beer soaked ramblings follow.
Man,

that's fast.​

Latex fast.​

that rhymes .....

 

[Steven G]

Example:

[math]\dfrac {x^7}{x^3} = \dfrac {x*x*x*x*x*x*x}{x*x*x*}= x*x*x*x = x^4[/math]. Do you see those 3 x's canceling out to 1?