# MATH TIME!

Discussion in 'User Topics' started by Mr.Lame, Jan 9, 2018.

1. ### Mr.LameVeteran

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BUT I do get it, so seperate for scientific notation.

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2. ### Mr.LameVeteran

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Perfect! Got the answer just before you posted • Like x 1
3. ### Mr.LameVeteran

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@MarkFL

(1/8)^-4/3

SO This time I didn't make a mistake on the question hahaha

do I make it a cube root ?

4. I would first use:

1/8 = 2^(-3)

like so:

(1/8)^(-4/3) = (2^(-3))^(-4/3)

And now multiply the exponents:

(1/8)^(-4/3) = (2^(-3))^(-4/3) = 2^4 = 16

5. ### Mr.LameVeteran

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A bit lost on the (2^(-3))^(-4/3)

6. 1/8 = 1/(2^3) = 2^(-3)

Is that the part that needed clarification?

7. ### Mr.LameVeteran

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1/8 = 2^(-3) < I don't understand how you get this any chance you could break that down.

From a previous I'd root with the denominator [(ex. 3/2) Square root (4)= 2^3)] and use the numerator as the exponent outside the root

8. ### Mr.LameVeteran

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So if I did this.

Cuberoot of 1/8 = 1/2^-4 = 16

9. We have:

1/8

Now, if we recognize that 8 is the cube of 2, that is 8 = 2^3, then we can write:

1/8 = 1/(2^3)

Now, when an exponent is down in the denominator, we can bring that term up into the numerator by negating the exponent

1/8 = 1/(2^3) = 2^(-3)

We can think of this as:

1/(2^3) = (2^0)/(2^3) = 2^(0 - 3) = 2^(-3)

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10. Yes, there's lots of ways to go about it...I try to show the method that I think will be the most useful in general. • Like x 1
11. ### Mr.LameVeteran

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I like that because there may be a situation where this method may not work!

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12. ### Mr.LameVeteran

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Hey @MarkFL,

When there are exponents inside a root, you divided by the root?

example 4root (x^8y^12) would be x^2y^3 right? Or am I doing it wrong?

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13. ### Mr.LameVeteran

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Thanks Mark , Getting better!!! • Love x 1
14. Just a note here...the simplification we stated only works for a non-negative value of y. • Winner x 1
15. ### Mr.LameVeteran

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O: Good to know, I'm sure I'm going to be seeing a negative one soon, so be prepared hahaha!

16. ### Mr.LameVeteran

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17. The short story here is that we started out with an even exponent on y and ended up with an odd exponent, and when a negative number has an odd exponent, the result is a negative value. But, when a negative number has an even exponent, the result is positive So, if y is negative, the original expression is positive, and we need to insure the end result is too. • Winner x 1