Question

Help solve!!!

the quantity of x to the four thirds power, over x to the five sixths power

the sixteenth root of the quantity of x times x to the third times x to the fourth

Answer 1

@Mertsj

Answer 2

@TuringTest

Answer 3

@zimmah

Answer 4

Is this the first problem? \[|\large \frac{ x^{4/3} }{ x^{5/6} }\]

Answer 5

yes

Answer 6

Use the identity: \[\large \frac{ x^m }{ x^n} = x^{m-n}\]

Answer 7

idk how

Answer 8

\[\large \frac{ x^{4/3} }{ x^{5/6} } = x^{4/3 - 5/6} = x^{8/6 - 5/6} = x^{3/6} = x^{1/2}\]

Answer 9

the second one is 16 on the outside of the radical symbol and x times x^3 times x^4
under the radical symbol

Answer 10

morganfaith__ keep in mind that -> \(\bf \Large{ a^{-{\color{red} n}} = \cfrac{1}{a^{\color{red} n}} \qquad \qquad

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

\cfrac{x^{\frac{4}{3}}}{x^{\frac{5}{6}}}\implies \cfrac{x^{\frac{4}{3}}}{1}\cdot \cfrac{1}{x^{\frac{5}{6}}}\implies \cfrac{x^{\frac{4}{3}}}{1}\cdot x^{-\frac{5}{6}}\\ \quad \\
\implies x^{\frac{4}{3}+(-\frac{5}{6})}\implies x^{\frac{4}{3}-\frac{5}{6}}}\)

Answer 11

what about the
second one

Answer 12

yes

Answer 13

x^3 times x^4 are both inside the radical and a small 16 is to the left of the radical?

Answer 14

\[\Large \sqrt[16]{x^3 \times x^4}?\]

Answer 15

but its x*x^3*x^4 under the radical symbol

Answer 16

\(\huge \bf \sqrt[16]{x\cdot x^3\cdot x^4}\quad ?\)

Answer 17

jdoe has it right

Answer 18

well, keep in mind that \(\Large \bf a^{\frac{{\color{blue} n}}{{\color{red} m}}} = \sqrt[{\color{red} m}]{a^{\color{blue} n}}\)

Answer 19

change all "radicand" to rational exponentials, and add away

Answer 20

to what?

Answer 21

or .. ahemm... well. just add inside the radicand, and then change to rational exponent

Answer 22

so it would be 3x^7?

Answer 23

of just x^7

Answer 24

\(\bf \large \sqrt[16]{x\cdot x^3\cdot x^4}\implies \sqrt[16]{x^1\cdot x^3\cdot x^4}\)

Answer 25

x * x^3 * x^4 = x^(1+3+4) = ?

Answer 26

8?

Answer 27

x^8

Taking the 16th root is same as raising to the exponent (1/16)

Answer 28

so the answer is x^8

Answer 29

No. (x^8)^(1/16) = ?

Answer 30

\(\Large \bf \sqrt[{\color{red}{ 16}}]{x\cdot x^3\cdot x^4}\implies \sqrt[{\color{red}{ 16}}]{x^1\cdot x^3\cdot x^4}\implies \sqrt[{\color{red}{ 16}}]{x^\square }\implies x^{\frac{\square }{{\color{red}{ 16}}}}\)

Answer 31

so x^8/16

Answer 32

simplify the exponent.

Answer 33

which is x^1/2?

Answer 34

Yes. The answers to both problem is x^(1/2) or square root of x.

Answer 35

thanks so much!!!

Answer 36

You are welcome.