Question

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Answer 1

Answer :

k = m/n

Solution :

$\text{Let } \sf log_{x^{n} }y^{m} = k log_{x}y = t$

Using Power rule : log a^m = m.log a, the equation becomes

$\implies\sf log_{x^{n} }y^{m} = log_{x}y ^{k} = t$

Simplying equations individually

$\implies\sf log_{x^{n} }y^{m} = t$

Writing it in exponential form [ If log_a N = x then a^x = N ]

$\implies\sf (x^{n})^{t} = y^{m}$

$\implies\sf (x^{t})^{n} = y^{m}$

$\implies\sf x^{t} = y^{ m/n}\longrightarrow eq(1)$

$\implies\sf log_{x}y ^{k} = t$

Writing it in exponential form

$\implies\sf x^{t} = y^{k}\longrightarrow eq(2)$

From eq( 1 ) and eq( 2 )

$\implies\sf y^{k} = y^{ m/n}$

Since bases are equal we can equate powers

$\implies\boxed{\sf k = \frac{m}{n} }$

Therefore the value of k is m/n.