Question

Need help with questions!

Answer 1

We always read our roots like this: \[\sqrt[2]{2} = 2^{1/2} \rightarrow \sqrt[3]{2} = 2^{1/3} \]

Answer 2

But in the case where it's a 2, we don't put the 2, it's just accepted that it's a 2

Answer 3

\[\sqrt[45]{5} = 5^{1/45}\]

Answer 4

etc.

Answer 5

\[\sqrt[3]{n ^{4}}= n ^{4/3}\]

Answer 6

Is one more example.

Answer 7

I know this is more than you need - but in general \[\sqrt[a]{c ^{b}} = c ^{b/a}\]

Answer 8

I still dont understand im sorry

Answer 9

\(\huge a^{\frac{{\color{blue} n}}{{\color{red} m}}} = \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad

\sqrt[{\color{red} m}]{a^{\color{blue} n}}=a^{\frac{{\color{blue} n}}{{\color{red} m}}}\)

Answer 10

Alright. So let's say you have the sqrt(4)

It looks like this: \[\sqrt{4}\]

Whenever we dont see a number inside the square root, we take it to be a 2. (Hence the name square)

So \[\sqrt{4}=\sqrt[2]{4}\]

It is saying the same thing.

But, when we take a cube root. Or a 4th root, they look like this: \[\sqrt[3]{4}\] or \[\sqrt[4]{4}\]

Answer 11

Oh! so for the first one the option would be the first one?

Answer 12

\(\huge \sqrt[10]{7}\implies \sqrt[{\color{red}{ 10}}]{7^{\color{blue}{ 1}}}\implies ?\)

Answer 13

All of that being said. A square root of 4 looks like \[\sqrt{4}\] Which is the same thing as \[\sqrt[2]{4}\]

Now you know that the square root of 4 is 2 right?

Well \[\sqrt{4} = \sqrt[2]{4} = 4^{1/2}\]

The 4 in this case is raised to the 1 power under the root.

\[4^{1} = 4 \rightarrow \sqrt{4} = 4^{1/2}\]

Answer 14

hmmm maybe I got the wrong one... but anyhow...

Answer 15

Anyways... with a square root of 4 they are not showing the 4 raised to the 1 power, or the 2 inside the root symbol but

\[\sqrt{4} \rightarrow \sqrt[2]{4^{1}} \rightarrow 4^{1/2}\]

Answer 16

Wow, that's a hard concept to explain actually. I hope you understand now lol

Answer 17

hmmm which one are you doing @nicolexcx ? did you follow Cosmichaotic lines?

Answer 18

Im trying to do the very first one but for the 10square root of 7 would it be 10 square root of 70?

Answer 19

@jdoe0001 Explained it perfect.

\[\sqrt[b]{m ^{a}} = m ^{a/b}\]

Answer 20

\[10\sqrt{7} = 10\times 7^{1/2}\]

Answer 21

\(\huge 5^{\frac{1}{2}}-\sqrt{5}?\)

Answer 22

hmmm ohhh \(\huge \sqrt[10]{7}?\)

Answer 23

\[10^{1/7}\]?

Answer 24

hmmm recall \(\huge { \bf a^{\frac{{\color{blue} n}}{{\color{red} m}}} = \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad

\sqrt[{\color{red} m}]{a^{\color{blue} n}}=a^{\frac{{\color{blue} n}}{{\color{red} m}}}

\\ \quad \\ \quad \\

\huge \sqrt[10]{7}\implies \sqrt[{\color{red}{ 10}}]{7^{\color{blue}{ 1}}}\implies ? }\)

Answer 25

\[7^{1/10}\]

Answer 26

yeap

Answer 27

and surely you can see what \(\Large \bf \sqrt[8]{5}\) is

Answer 28

I understand that now.. But i do still need help with the first 2

Answer 29

hmmm which one \(\huge 5^{\frac{1}{2}}-\sqrt{5}?\)

Answer 30

yes it would be \[5^{1/8}\]

Answer 31

for \(\Large \bf \sqrt[8]{5}\) yeap

Answer 32

yes! okay well how about this one?

Answer 33

well \(\Large \bf 5^{\frac{{\color{blue}{ 1}}}{{\color{red}{ 2}}}}-\sqrt{5}\implies \sqrt[{\color{red}{ 2}}]{5^{\color{blue}{ 1}}}-\sqrt{5}\implies \sqrt{5}-\sqrt{5}\implies ?\)

Answer 34

notice, the left side looks like the right side
thus

\(\bf same - same = ?\)

Answer 35

ohhh wait a sec... shoot got a typo there. .is supposed to be 15 lemme fix that quick

Answer 36

so \(\large \bf 15^{\frac{{\color{blue}{ 1}}}{{\color{red}{ 2}}}}-\sqrt{15}\implies \sqrt[{\color{red}{ 2}}]{15^{\color{blue}{ 1}}}-\sqrt{15}\implies \sqrt{15}-\sqrt{15}\implies ?\)

Answer 37

notice, the left side looks like the right side

Answer 38

so \[15^{1/2}\]

Answer 39

well... yes.... you can rewrite it that way too
you'd still be ending up with " same - same "

\(\Large \bf 15^{\frac{1}{2}}-\sqrt{5}\implies 15^{\frac{1}{2}}-15^{\frac{1}{2}}\implies ?\)

Answer 40

\[\sqrt{5}\]

Answer 41

what do you get when you subtract say \(\Large \bf same - same =?\)

Answer 42

0? or the same?

Answer 43

yeap \(\large \bf 15^{\frac{{\color{blue}{ 1}}}{{\color{red}{ 2}}}}-\sqrt{15}\implies \sqrt[{\color{red}{ 2}}]{15^{\color{blue}{ 1}}}-\sqrt{15}\implies \cancel{ \sqrt{15} }-\cancel{ \sqrt{15} }\implies 0\)

Answer 44

ohhhhhh! so then what would my option be?

Answer 45

hmm

Answer 46

is that a - or an equals?

Answer 47

I gather is an equals

Answer 48

=

Answer 49

yeah its equala

Answer 50

equals

Answer 51

I don't see any of those choices well fit to provide a proof for it

Answer 52

Im assuming you dont see it for this one either

Answer 53

well... I don't see any proof there.... thus

all they're doing is raising it to 2 or getting its reciprocal or both
but that only yields some other valid value, but is not proof of it

Answer 54

if you dont can you explain to me why \[9^{1/12}\] is equivalent to \[1/12\sqrt{9}\] ?

Answer 55

hopefully this makes more sense

Answer 56

hmmmm not sure what they're expecting

Answer 57

I do know that \(\bf \large {
a^{-\frac{{\color{blue} n}}{{\color{red} m}}} =
\cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad thus
\\ \quad \\
9^{-\frac{1}{12}}\implies \cfrac{1}{9^{\frac{1}{12}}}\implies \cfrac{1}{\sqrt[12]{9}}
}\)

Answer 58

okay what about these 2

Answer 59

I can do the same procedure... but I'm not sure is what is being expected though

Answer 60

please! it would help me understand it better

Answer 61

actually its okay! thank you anyways!

Answer 62

hmm ok :)