Question

If log(x + y) = log 3 + ½log x + ½log y, prove that x² + y² = 7xy​

Answer 1

Answer:

See Explanation

Step-by-step explanation:

$log(x + y) = log3 + \frac{1}{2} logx+ \frac{1}{2} logy \\ \\ log(x + y) = log3 + logx ^{\frac{1}{2}} + logy ^{\frac{1}{2}}\\ \\ log(x + y) = log3 + log(xy) ^{\frac{1}{2}} \\ \\ log(x + y) = log[3(xy) ^{\frac{1}{2}}] \\ \\ x + y = 3(xy) ^{\frac{1}{2}} \\ \\ squaring \: both \: sides \\ {(x + y)}^{2} = \bigg(3(xy) ^{\frac{1}{2}} \bigg)^{2} \\ \\ {x}^{2} + {y}^{2} + 2xy = 9xy \\ \\ {x}^{2} + {y}^{2} = 9xy - 2xy \\ \\ \purple{ \bold{{x}^{2} + {y}^{2} = 7xy}} \\ thus \: proved$

Answer 2

Answer:

$\boldsymbol {Rule} : \\\\ \boldsymbol {\log _ n a+\log _n b = \log _n a\cdot b} \\\\ \boldsymbol {n \cdot \log _a b =\log_a b ^n } \\\\ \large \boldsymbol {} \log _n (x+y)= \log_n 3 + \frac{1}{2} \log_n x+\frac{1}{2} \log _n y \\\\ \log _n (x+y) = \log _n 3+\frac{1}{2} (\log_n xy ) \\\\ \log _n ( x+y ) = \log_n 3+\log _n \sqrt{xy } \\\\ \log _n (x+y) = \log_n \ 3 \sqrt{xy } \\\\ x+y= 3\sqrt{xy} \\\\ (x+y) ^2= (3\sqrt{xy} })^2 \\\\ x^2+y^2+2xy =9 xy \\\\ \boldsymbol {x^2+y^2=7xy }$