Question

if a^p=b^q=c^r=d^s and ab=cd then the value of 1/p+1/q-1/r-1/s reduces to

Answer 1

Answer:

0

Step-by-step explanation:

Let a^p=b^q=c^r=d^s = k

therefore, a= k^(1/p)

                 b= k^(1/q)

                 c= k^(1/r)

                 d= k^(1/s)

we know that ab = cd,

therefore, k^(1/p)*k^(1/q) = k^(1/r)*k^(1/s)

            => (1/p)+(1/q) = (1/r)+(1/s)                                   { since bases are equal }

            => (1/p)+(1/q)-(1/r)-(1/s) = 0

Answer 2

a^p = b^ q

a= b^ ( q/p)

b^ q= c^ r

c= b^ ( q/r)

b^q = d^ s

d= b^ ( q/s)

As ab = cd

b^ ( q/p) b = b^ ( q/r) b^ ( q/s)

b^ ( q/p +1) = b^ ( q/r + q/s)

compare

q/p+ 1 = q/r + q/s

q/p - q/r - q/s + 1 = 0

q/p - q/r - q/s + q/q = 0

1/p - 1/r - 1/s + 1/q = 0