Simplifying radicals(algebra II)

[liz777]

Homework Statement

(cube root $$\sqrt[]{}x^4$$ * $$\sqrt[]{}x^5$$)^-2

Honestly, I tried and my answer didnt make any sense...The answer is supposed to be 1/(x^23/3)

How would I go about solving this? Any help would be appreciated :)

Also how would I solve the cube root of 7 * the cube root of 49?

 

[Mark44]

What you wrote is very difficult to decipher. How about rewriting it using exponents rather than radicals? The cube root of a is a^(1/3).

The cube root of 7 times the cube root of 49 is the cube root of 343, which can be simplified.

 

[v0id19]

Also how would I solve the cube root of 7 * the cube root of 49?

Think about what you have: 2 cube roots. One is $$\sqrt[3]{7}$$ and the other is $$\sqrt[3]{7^2}$$. What can you do with that?

 

[liz777]

ok I get that one! you would get cube root of 7^3...so you would get 7. thanks I get that one now!

and back to the first problem. in exponents I think it would be:

(X^3/4 * X^1/5)^-2

So if anyone wants to help with this one...so far I still have the wrong answer, I have no idea what I'm doing wrong!

 

[symbolipoint]

Laws of Exponents are taught in Algebra 1 and Algebra 2.

a^-m=$$\frac{1}{a^m}$$
That one did not type set correctly. I'm trying to state
'a' to the negative m power equals the fraction one over a to the m power. a^(-m)=1/(a^m)

a^maⁿ=a^m+n

 

[Mark44]

ok I get that one! you would get cube root of 7^3...so you would get 7. thanks I get that one now!

and back to the first problem. in exponents I think it would be:

(X^3/4 * X^1/5)^-2

So if anyone wants to help with this one...so far I still have the wrong answer, I have no idea what I'm doing wrong!

That doesn't look like what you started with, which I think should look more like this:
(x^4/3 * something)^-2

The first factor inside the parentheses seems to be the cube root of x^4, which is x^4/3. The other factor appears to be the square root of x^5. Can you tell us exactly what the problem is that you're working on?