Question

If x square+y square=7xy then show that 2log(x+y)=logx+logy+2log3

Answer 1

$\underline{\underline{\Large{\mathfrak{Solution :}}}}$



$\underline{\mathsf{Given \longrightarrow x^2 \:+ \: y^2 \: = \: 7xy }}$



$\textsf{Add 2xy to both sides : }$



$\sf \implies x^2 \: + \: y^2 \: + \: 2xy \: = \: 7xy \: + \: 2xy \\ \\ \sf \implies (x \: + \: y )^2 \: = \: 9xy$



$\textsf{Applying Logarithm : }$



$\sf \implies log \: (x \: + \: y)^{2} \: = \: log \: 9xy \\ \\ \\ \sf \implies 2 \: log \: (x \: + \: y) \: = \: log \: ( 9 \: \cdot \:x \cdot y) \\ \\ \\ \sf \implies 2 \:log \: (x \: + \: y) \: = \: log \: 9 \: + \: log \: x \: + \: log \: y \\ \\ \\ \sf \implies 2 \: log \: (x \: + \: y) \: = \: log \: {3}^{2} \: + \: log \: x \: + \: log \: y \\ \\ \\ \sf \implies 2 \: log \: (x \: + \: y) \: = \: 2 \: log \: 3 \: + \: log \: x \: + \: log \: y \\ \\ \\ \sf \therefore \quad \: 2 \: log \: (x \: + \: y) \: = \: log \: x \: + \: log \: y \: + \: 2 \: log \: 3$



$\underline{\textsf{Algebraic identity used : }} \\ \\ \sf \implies a^2 \: + \: b^2 \: + \: 2ab \: = \: ( a \: + \: b )^2$



$\underline{\textsf{Logarithm rules used : }} \\ \\ \mathsf{ \implies log_{a} \: bc \: = \: log_{a} \: b \: + \: log_{a} \: c} \\ \\ \mathsf{ \implies log_{a} \: b^c \: = \: c \: log_{a} \: b }$

Answer 2

LHS=2log(x+y)
$= log( {x + y)}^{2} \\ = log( {x}^{2} + + {y}^{2} + 2xy ) \\ = log(7xy + 2xy) \\ = log(9xy) \\ = logx + logy + log9 \\ = logx + logy + log {3}^{2} \\ = logx + logy + 2 log3$