Question

Prove this : logarithms ​

Answer 1

$\underline{\textbf{\large{ Given :}}}$

$log( \frac{x + y }{3} ) = \frac{1}{2} logx + \frac{1}{2} logy$

$\underline{\textbf{\large{ To prove :}}}$

$\frac{x}{y} + \frac{y}{x} = 7$

$\underline{\textbf{\large{ proof: }}}$

$log( \frac{x + y}{3} ) = \frac{1}{2}logx + \frac{1}{2} logy$

→ $log( \frac{x + y}{3} ) = \frac{1}{2}(logx + logy)$

▶ Apply log rule[ loga + logb = logab ]

→ $log( \frac{x + y}{3} ) = \frac{1}{2}logxy$

▶ Apply log rule [log m^(n) = n log m ]

→ $log( \frac{x + y}{3} ) = log {(xy)}^{ \frac{1}{2} }$

→ $( \frac{x + y}{3} ) = {(xy)}^{ \frac{1}{2} }$

Squaring on both the sides,

→ ${ (\frac{x + y}{3} )}^{2} = { ({(xy)}^{ \frac{1}{2} }) }^{2}$

→ $\frac{ {(x + y)}^{2} }{9} = xy$

→ ${(x + y)}^{2} = 9xy$

→ ${x}^{2} + 2xy + {y}^{2} = 9xy$

→ ${x}^{2} + {y}^{2} = 9xy - 2xy$

→ ${x}^{2} + {y}^{2} = 7xy$

Dividing xy on both the sides

→ ${ \frac{ {x}^{2} + {y}^{2} }{xy} } = 7$

→ $\frac{ {x}^{2} }{xy} + \frac{ {y}^{2} }{xy} = 7$

→ $\frac{x}{y} + \frac{y}{x} = 7$

$\underline{\textbf{hence proved}}$