Question

solve

Answer 1

1st one:
cube root of 64 is 4, cube root of 27 is 3. Ans=4/3

Answer 2

Simplified 2nd one to x^8=3,814,697,265,625-don't know where to go from there.

Answer 3

No idea 4 the rest

Answer 4

For the second one, you don't need to exponentiate. Simply recognize that the 18th root of something to the 18th power is that number (but it's the positive version of that number)

Basically, \[\sqrt[18]{x^{18}}=|x|\]

Answer 5

oops got cut off, but you keep going and you'll basically get positive 5 as your answer for #2

Answer 6

do you know what to do for the other ones?
I didn't understand some of the terms panther was using

Answer 7

sure, the first one is solved as follows

\[\sqrt[3]{\frac{64}{27}}\]

\[\frac{\sqrt[3]{64}}{\sqrt[3]{27}}\]

\[\frac{\sqrt[3]{4^3}}{\sqrt[3]{3^3}}\]

\[\frac{4}{3}\]


So \[\sqrt[3]{\frac{64}{27}}=\frac{4}{3}\]

Answer 8

What about the last two?

Answer 9

Basically, the cube root of something cubed, is just that something (odd terminology, but it works). Eg: 2 cubed is 8. The cubed root of 8 is 2.

Answer 10

Right. I understand that question, and i get the second one too now, but I don't understand the last two questions.
What are rational exponents and radical notation?

Answer 11

Third problem

\[\sqrt[2]{7}\sqrt[4]{3}\]

\[7^{\frac{1}{2}}\cdot3^{\frac{1}{4}}\]

\[7^{\frac{2}{4}}\cdot3^{\frac{1}{4}}\]

\[\left(7^{2}\cdot3^{1}\right)^{\frac{1}{4}}\]

\[\left(49\cdot3\right)^{\frac{1}{4}}\]

\[\left(147\right)^{\frac{1}{4}}\]

\[147^{\frac{1}{4}}\]

\[\sqrt[4]{147}\]


So after all that, \[\sqrt[2]{7}\sqrt[4]{3}=\sqrt[4]{147}\]

Answer 12

Rational exponents are exponents that are fractions. They represent the idea that the expression involves radicals.

Answer 13

oh. So if you take the square root of something, you're really putting it to the 1/2 power?

Answer 14

Last question (#4)

\[16^{\frac{3}{4}}\]

\[\left(16^{\frac{1}{4}}\right)^{3}\]

\[\left(\sqrt[4]{16}\right)^{3}\]

\[\left(\sqrt[4]{2^4}\right)^{3}\]

\[\left(2\right)^{3}\]


\[8\]


So \[16^{\frac{3}{4}}=8\]

Answer 15

Yes, taking the square root of any number is the same as raising that number to the 1/2 power. Eg: \[\sqrt{2}=2^{\frac{1}{2}}\]

Answer 16

Okay I don't understand one thing you're doing- it looks like you're factoring out the exponents? I didn't know you could do that- or, at least, I've never seen it done before... but it seems to serve you well.

Answer 17

which one are you referring to?

Answer 18

all of them! It's just a process that I haven't seen yet. What are the can and cannot do's for that? Do you factor them like normal numbers?

Answer 19

Basically, I'm using the idea that

\[\left(x^y\right)^{z}=x^{y\cdot z}\]

Essentially this says that you can multiply powers if you raise an exponential expression to another exponent. So I can then use the properties of multiplication (factoring, rearranging, etc) to simplify. Hopefully that clears things up a bit.

Answer 20

Oh and I'm also using the idea that

\[\sqrt[n]{x^m}=x^{\frac{m}{n}}\]


which helps me convert to and from radical and exponential notation.

Answer 21

ok. thanks for explaining