Another College Algebra problem......

[colepoin]

I'm not sure what's the best way to enter this problem, but here goes...


(2x^1/2)^6

divided by


(4x^2/3)^3



*Each ^ indicates an exponent.
* Anywhere there is a / it means it is a fraction.


Thanks if you can help me out!

 

[nyc_function]

Exponents....

I'm not sure what's the best way to enter this problem, but here goes...


(2x^1/2)^6

divided by


(4x^2/3)^3



*Each ^ indicates an exponent.
* Anywhere there is a / it means it is a fraction.


Thanks if you can help me out!

(2x^1/2)^6/(4x^2/3)^3

Do the numerator first.

Raise every term in the parentheses to the 6th power.

2^6 = 32

x^(1/2)^6 = x^3

In the numerator we have 32x^3.

In the denominator, raise every term to the 3rd power.

(4x^2/3)^3


4^3 = 64


x^(2/3)^3 = x^2


In the denominator we have 64x^2.


Together we now have a new fraction:


(32x^3)/(64x^2) = x/2 or (1/2) times (x)...either one is correct.


Does this make sense?

 

[colepoin]

Still not sure how to finish that!

 

[lookagain]

(2x^1/2)^6/(4x^2/3)^3

No, JeffM showed the correct form, and it is because
of the Order of Operations.

Do the numerator first.

Raise every term in the parentheses to the 6th power.

2^6 = 32

[x^(1/2)]^6 = x^3
Eliminate the ambiguity of the exponents. **

In the numerator we have 32x^3.

In the denominator, raise every term to the 3rd power.

(4x^2/3)^3
The fraction here must be in grouping symbols.


4^3 = 64


[x^(2/3)]^3 = x^2 See ** above.


In the denominator we have 64x^2.


Together we now have a new fraction:


32x^3/64x^2 =
Although, this may be mathematically correct, it would
be clearer more quickly as (32x^3)/(64x^2).

.