Question

if p^1/x =q^1/y=r^1/z and pqr=1, then prove that x+y+z=0.​

Answer 1

Question:

If p^(1/x) = q^(1/y) = r^(1/z) and p•q•r = 1, then prove that x + y + z = 0.

Note:

• (a^m)×(a^n) = a^(m+n)

• (a^m)/(a^n) = a^(m-n)

• (a^m)×(b^m) = (a×b)^m

• (a^m)/(b^m) = (a/b)^m

• [a^m]^n = a^(m×n)

• a^0 = 1

• If a^m = a^n then m = n

• If a^m = b then a = b^(1/m)

Solution:

Given:

p^(1/x) = q^(1/y) = r^(1/z)

p•q•r = 1

To prove:

x + y + z = 0

Proof:

Let p^(1/x) = q^(1/y) = r^(1/z) = k

Then ,

=> p^(1/x) = k

=> p = k^x --------(1)

Similarly,

=> q^(1/y) = k

=> q = k^y --------(2)

Similarly,

=> r^(1/z) = k

=> r = k^z ---------(3)

Now,

Multiplying eq-(1) , (2) and (3) , we have ;

=> p•q•r = (k^x)•(k^y)•(k^z)

=> p•q•r = k^(x+y+z)

=> 1 = k^(x+y+z) { Given : p•q•r = 1 }

=> k^0 = k^(x+y+z)

=> 0 = x + y + z

=> x + y + z = 0

Hence proved.

Answer 2

Question :- if p^(1/x) = q^(1/y) = r^(1/z) , and pqr = 1 , prove that, x + y + z = 0 .

Solution :-

Let us assume that,

→ p^(1/x) = q^(1/y) = r^(1/z) = k , (where k is a constant number.)

then,

→ p^(1/x) = k

→ p = k^x

similarly,

→ q^(1/y) = k

→ q = k^y

and,

→ r^(1/z) = k

→ r = k^z .

given that,

→ p * q * r = 1 .

putting above values we get,

→ k^x * k^y * k^z = 1

using a^l * a^m * a^n = a^(l + m + n) , we get,

→ k^(x + y + z) = 1

now, we know that, if power of a number is 0 , it is equal to 1,

therefore,

→ k^(x + y + z) = k^0

now, we know that, when base is same, powers are equal . { a^m = a^n => m = n }

hence,

→ (x + y + z) = 0 (Proved.)

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