Having a hard time computing with the ^ sign?

Question

(p^-5)^-9 (p^-6)^4
______________
(p^-8)^0 (p^4)^-5

Answer 1

using

(a^b)^c = a^(bc) you get numerator = p^45 * p^ -24
denominator = p^0.p^-20

using (x^a)*(x^b) = x^(a+b) numerator = p^21 denominator = p^-20

so product = p^41 using p^a/p^b = p^(a-b)

Answer 2

Follow exponent laws:

(x^m)(x^n)=x^m+n

X^m/X^n=X^m-n

(X^m)^n=X^mn

(xy)^m=x^m*y^m

(x/y)^n=x^n/y^n

X^-n=1/x^n

(x/y)^-n=(y/x)^n

X^1=x

X^0=1


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Numerator: (p^-5)^-9=P^-5*-9=p^45
Numerator 2nd: (p^-6)^4=p^-6*4=p^-24

Numerator Multiplication: (p^45)(p^-24)=p^45+(-24)=p^21

Denominator: (p^-8)^0=P^-8*0=P^0=1(hind anything to the power of 0 is automatically 1)
Denominator 2nd: (p^4)^-5=p^4*-5=p^-20

Denominator Multiplication(it is not always multiplication but in this particular question since there is no +, - seperation then it means it is multiplication):
(p^0)(p^-20)=p^0+(-20)=p^-20


then p^21/p^-20= p^21-(-20)=

Answer to the question:

=================p^41

see it is quite simple upon following the exponent laws and doing it in a clear methodical system as such. Make numerator into single variable, and also denominator, and then simply subtract the bottom exponent from the top one.

Answer 3

(p^-5)^-9 (p^-6)^4
______________
(p^-8)^0 (p^4)^-5


uhh ... numerator first ..
(p^-5)^-9 (p^-6)^4= (p^ 45) * (p^-24) = p^21


denominator:
(p^-8)^0 (p^4)^-5 = 1 * p^-20

p^21 / p^(-20) = p ^ 41


unless I made a mistake ...

Answer 4

((p^-5)^-9 * (p^-6)^4) / ((p^-8)^0 * (p^4)^-5)

(p^(-5 * -9) * p^(-6 * 4)) / (p^(-8 * 0) * p^(4 * -5))

(p^45 * p^-24)/(p^0 * p^-20)

(p^(45 + (-24)))/(1 * p^-20)

(p^(45 - 24)) / (p^-20)

(p^(21)) / (p^(-20))

p^(21 - (-20))

p^(21 + 20)

ANS : p^41

Answer 5

When you exponentiate, you multiple exponents.

(p^-5)^-9=p^45
(p^-6)^4=p^-24
(p^-8)^0=p^0=1
(p^4)^-5=p^-20
so
p^45*p^-24/p^-20 when you multiply, add exponents, when you divide subtract so
p^(45-24-(-20))=p^41