Question

1. If x2 + y2 = 7xy, prove that log (x + y) = log 3 + 1/2(logx+log y)​

Answer 1

Answer:

Given x² + y² = 7xy

=> x² + y² + 2xy = 7xy + 2xy

=> (x+y)² = 9xy

=> log(x+y)² = log (3²xy)

=> 2log(x+y) = log3² + logx + logy

$log a^{m} = m loga$

$log (mn) = log m + log n$

=> 2log(x+y)= 2log3 + log x + log y

/* divide both sides by 2, we get

$log(x+y) =log3 +\frac{1}{2} (logx+logy)$

•••♪

Answer 2

Step-by-step explanation:

Given x² + y² = 7xy

=> x² + y² + 2xy = 7xy + 2xy

=> (x+y)² = 9xy

=> log(x+y)² = log (3²xy)

=> 2log(x+y) = log3² + logx + logy

log a^{m} = m logaloga

m

=mloga

log (mn) = log m + log nlog(mn)=logm+logn

=> 2log(x+y)= 2log3 + log x + log y

/* divide both sides by 2, we get

log(x+y) =log3 +\frac{1}{2} (logx+logy)log(x+y)=log3+

2

1

(logx+logy)