Question

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Answer 1

QUESTION :

$\: \: \: \: \: \bigstar \: \: \: \sf{ \large{If \: \: p {}^{\frac{1}{x}} = q{}^{\frac{1}{y}} = r {}^{\frac{1}{z}} \: \: and \: \: pqr = 1, \: prove \: }} \\ \sf{ \large{that \: \: x + y + z = 0}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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IMPORTANT IDENTITIES :

$1. \: \: \boxed{ \rm{a {}^{m} \times a {}^{n} = a {}^{m + n}}} \: \: \: \: \: \: \: \: \: \: \\ \\ 2. \: \: \boxed{ \rm{a {}^{0} = 1}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: 3. \: \: \boxed{ \rm{If \: \: a {}^{m} = b {}^{m}, \: then \: \: m = n }}\\ \\ \: \: \: \: \: \: 4. \: \: \boxed{ \rm{If \: \: a {}^{ \frac{1}{m} } = b, \: then \: \: a = b {}^m}}$

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SOLUTION :

❒ Given :-

• $\: \bigstar \: \: \sf{p {}^{\frac{1}{x}} = q{}^{\frac{1}{y}} = r {}^{\frac{1}{z}}}$

• $\: \bigstar \: \: \sf{pqr = 1}$

❒ To Prove :-

• $\: \bigstar \: \: \sf{x + y + z = 0}$

❒ Proof :-

Let,

• $\bigstar \: \: \sf{ p{}^{\frac{1}{x}} = q{}^{\frac{1}{y}} = r {}^{\frac{1}{z}} = k}$

Then,

$\sf{ \therefore \: \: p{}^{\frac{1}{x}} =k} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{\implies \: p = k {}^{x} \: \: \: [Using \: \: identiy \: \: (4)] \: - - - > (i)}$

$\sf{\therefore \: \: q{}^{\frac{1}{y}} =k} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{\implies \: q = k {}^{y} \: \: \: [Using \: \: identiy \: \: (4)] \: - - - > (ii)}$

$\sf{\therefore \: \: r{}^{\frac{1}{z}} =k} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{\implies \: r = k {}^{z} \: \: \: [Using \: \: identiy \: \: (4)] \: - - - > (iii)}$

Multiplying eq. (i), eq. (ii) and eq. (iii), we get :-

$\bigstar \: \: \: \sf{p \times q \times r = k {}^{x} \times k {}^{y} \times k {}^{z}} \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: pqr = k {}^{(x + y + z)} \: \: [Using \: \: identiy \: \: (1)]} \\ \\ \sf{ \implies \: 1 = k {}^{(x + y + z)} \: \: \: \: \: [Given]}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: k {}^{0 } =k {}^{(x + y + z)} \: \: \:[Using \: \: identiy \: \: (2)]} \: \\ \\ \sf{ \implies \: \: 0 = x + y + z \: \: [Using \: \: identiy \: \: (3)]} \\ \\ \sf{ \therefore \: \: \underline{ \: x + y + z = 0 \: }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• ★ Hence, proved.

Answer 2

Correct 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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