Question

log(x^3-y^3/7)=log(x-y)+logx+logy . then find x/y+y/x
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Answer 1

Solution :-

$\rm log \bigg( \dfrac{ {x}^{3} - {y}^{3} }{7} \bigg) =log(x - y) + logx + logy$

$\implies \rm log \bigg( \dfrac{ {x}^{3} - {y}^{3} }{7} \bigg) =log(x - y) xy$

[ Because, by product rule, log a + log b = log ab ]

Comparing on both sides we get

$\implies \rm \dfrac{ {x}^{3} - {y}^{3} }{7} =(x - y) xy$

$\implies \rm \dfrac{(x - y)( {x}^{2} + xy + {y}^{2}) }{7xy} =(x - y)$

[ Because x³ - y³ = (x - y) (x² + xy + y²) ]

$\implies \rm \dfrac{ {x}^{2} + xy + {y}^{2} }{7xy} = \dfrac{x - y}{x - y}$

$\implies \rm \dfrac{ {x}^{2} + xy + {y}^{2} }{7xy} = 1$

$\implies \rm {x}^{2} + xy + {y}^{2} = 7xy$

$\implies \rm {x}^{2} + {y}^{2} = 7xy - xy$

$\implies \rm {x}^{2} + {y}^{2} = 6xy$

Dividing throughout by xy

$\implies \rm \dfrac{ {x}^{2} }{xy} + \dfrac{ {y}^{2} }{xy} = \dfrac{6xy}{xy}$

$\implies \rm \dfrac{x}{y} + \dfrac{y}{x} = 6$

Therefore the value of x/y + y/x is 6.