Powers w/Rational Exponents: Evaluate (Review My Work)

[calcdummy]

Homework Statement

Write as a single power, then evaluate:
a) (-32)^3/5 x (-32)^-4/5 / (-32)^2/5


b) 4096^3/6 / 4096^2/3 x 4096^5/6

Homework Equations

The Attempt at a Solution

a) (-32)^3/5 x (-32)^-4/5 / (-32)^2/5
= (-32)^3/5+(-4/5)-2/5
= -32^-3/5
= -1/8 <- not sure about this

b) 4096^3/6 / 4096^2/3 x 4096^5/6
= 4096^(9-8+10)/12
= 4096^11/12

I'm not so sure of where to go from here.

 

[emailanmol]

Your first answer is right :-)

Second needs some review.

Remember (x^a*x^b)/(x^c*x^d)

Is x^(a+b-c-d)

So its 4096^(3/6-2/3-5/6)


You have made a mistake by writing 4096^({9}-8-10)/12

It shouldn't be 9 in the curly brackets i put{} :-).

Also factorise 4096.
Write it in power of primes.

For eg 400=2^4*5^2

So (400)^(1/2)

Is
[(2^4)*(5^2)]^(1/2)

So its (2^2)*(5) which gives 20.

 

[calcdummy]

Oh man I copied down the wrong problem. I'm so sorry. It was supposed to be:
4096^3/4 / 4096^2/3 x 4096^5/6
I got the common denominator which would been 12. That is how I got 4096^(9-8+10)/12
= 4096^11/12

am I still incorrect?

 

[Dick]

Oh man I copied down the wrong problem. I'm so sorry. It was supposed to be:
4096^3/4 / 4096^2/3 x 4096^5/6
I got the common denominator which would been 12. That is how I got 4096^(9-8+10)/12
= 4096^11/12

am I still incorrect?

You should probably use parentheses instead of spaces to make that clearer. If you mean (4096^(3/4)/4096^(2/3))*4096^(5/6) then 4096^(11/12) is correct. There's a much simpler way to express that answer. 4096^(1/12)=2.