maths help, how to solve this?

Question

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

it is askig for to mention the above in " positive "times" "

(i'm not good in english terms. not sure what you call the number that comes after sign ^. in my language the number coomes after the sign ^ is called "darshaka", now it is asking to mention this in positive "darshaka", sorry if you got confused with my language. i'm not good in english terms.)

please try to understand and give me a good anser. it is bit urgent. thanks alot.

Answer 1

Darshana C
Maths help, how to solve this?

First, the result:

( (x^-2 y / x^3 y^-4) / (x y^-1 / x^-3 y^2) )^6 = y^48 / x^54.

The quantity following the ' ^ ' sign is called an EXPONENT, or less formally, a "power." The same terms are still used when the exponent is more complicated; thus y^(x - ct) means "raise y to the power (x - ct)."

Your question is asking you to simplify the given expression, writing it in a final form using only positive exponents. That means that you need to replace terms like x^(- 2) by 1 / x^2, and 1 / y^(- 4) by y^4.

First, I like to alternate brackets and parentheses, for greater clarity. [That way, I don't get confused by adjacent, doubled limiters like ' )) ' ! ]So I'll re-write your original expression slightly. It's:

[ (x^-2 y / x^3 y^-4) / (x y^-1 / x^-3 y^2) ]^6.

My approach after that is to start writing everything inside the (...) parentheses in terms of positive exponents. That gives:

[(y^5 / x^5) / (x^4 / y^3)]^6 = [y^8 / x^9]^6 = y^48 / x^54.

Live long and prosper.

Answer 2

Your darshaka = our exponent or power.
For example, x^-5 = 1/ x^5 and 1/x^-5 = x^5
( (x^-2 y / x^3 y^-4) / (x y^-1 / x^-3 y^2) )^6
= ((y/x^2/x^3/y^4)/(x/y /y^2/x^3))^6
= ((y^5/x^5)/(x^4/y^3))^6
= (y^8/x^9)^6
=y^48/x^54

Answer 3

((y^5/x^5)/(x^4/y^3))^6 --> ((y^8)/(x^9))^6