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Answer 1

Answer:

it equals to 1

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Answer 2

AnswEr :

• Option (B) is correct

Given Equations,

$\sf \: {p}^{x - 1} = qr - - - - - - (1) \\ \sf \: {q}^{y - 1} = rp \\ \sf \: r {}^{z - 1} = pq$

To finD

$\sf \: \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$

Consider Equation (1),

$\implies \: \sf \: \dfrac{ {p}^{x} }{p} = qr \\ \\ \implies \sf \: p {}^{x} = pqr$

Similarly,

$\sf \: {q}^{y} = pqr \: and \: {r}^{z} = pqr$

Therefore,

$\sf \: {p}^{x} = {q}^{y} = \: {r}^{z} = pqr$

Taking log in the expression,we get :

$\implies \: \sf \: log({p}^{x} ) =log( {q}^{y}) = log( {r}^{z} ) = log(pqr)$

Applying Power Rule of Logarithm,

$\implies \: \sf \: x.log({p}^{} ) =y.log( {q}^{}) = z.log( {r}^{} ) = log(pqr)$

We get,

$\sf \: x = \dfrac{log(pqr) }{log(p)} \\ \\ \implies \sf \: log(p) = \dfrac{log(pqr) }{x} - - - - - - - - - (a)$

Also,

$\sf \: y = \dfrac{log(pqr) }{log(q)} \\ \\ \implies \sf \: log(q) = \dfrac{log(pqr) }{y} - - - - - - - - - (b)$

And,

$\sf \: z = \dfrac{log(pqr) }{log(r)} \\ \\ \implies \sf \: log(r) = \dfrac{log(pqr) }{z} - - - - - - - - - (c)$

Adding equations (a),(b) and (c),we get :

$\longrightarrow \sf log(pqr) \bigg[\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg] = log(p) + log(q) + log(r)$

By Product Rule of Logarithm,

$\longrightarrow \sf \cancel{log(pqr)} \bigg[\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \bigg] = \cancel{log(pqr)}$

Thus,

$\longrightarrow \underline{\boxed{\sf \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}=1}}$