How do you write #(x^-2/y^-3)^-2#using only positive exponents?

Answer 1

# (x^2/y^3)^2 # or # x^4/y^6 #

Use #A^-1 = 1/A#

So # (x^-2/y^-3)^-2 = (y^-3/x^-2)^2 # # :. (x^-2/y^-3)^-2 = ((1/y^3)/(1/x^2))^2 # # :. (x^-2/y^-3)^-2 = ((1/y^3)x^2)^2 # # :. (x^-2/y^-3)^-2 = (x^2/y^3)^2 # or # x^4/y^6 #

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Answer 2

Eva Adamson

Numerous indices laws exist.

I believe the following will be the simplest to implement here:

#(xy)^m= x^my^m" and "(x^2 y^3)^4 = x^8y^12#

#(x^-2/y^-3)^-2" "larr# neg x neg = pos

=#x^4/y^6#

OR: Remember that there are two additional laws of indices.

#x^-m = 1/x^m" and "(x^m/y^n)^color(red)(-p) = (y^n/x^m)^color(red)((+p))#

#(x^-2/y^-3)^color(red)(-2) = (y^-3/x^-2)^color(red)(+2)#

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

=#x^4/y^6#

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Answer 3

Jace Anderson

To write (x^-2/y^-3)^-2 using only positive exponents, you apply the negative exponent rule, which states that a negative exponent indicates the reciprocal of the base raised to the positive exponent. Therefore, (x^-2/y^-3)^-2 becomes (y^3/x^2)^2, resulting in y^(32)/x^(22), which simplifies to y^6/x^4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.