Question

if x2+y2=34xy,then show that 2log (x+y)=2log 2+2log3+ log x+log y​

Answer 1

Answer:

Step-by-step explanation:

x² + y² = 34xy

Add 2xy both sides of the equation,

x² + y² + 2xy = 34xy + 2xy

( x + y )² = 36xy

( x + y )² = 6²xy

Take logarithm both sides ,

log ( x + y )² = log [ 6²xy ]

2log ( x + y ) = log 6²+log x + log y

[ Since 1 ) log a^n = n log a ,

2 ) log ab = log a + log b ]

2 log ( x + y ) = 2log 6 + log x + logy

2log(x + y) = 2log(2 × 3)+logx+logy

2log(x + y) = 2(log2 +log3)+logx+logy

2log(x+y)=2log2+2log3+logx+logy

Hence proved.

I hope this helps you.

Answer 2

Answer:

Hello ✌️

Step-by-step explanation:

➡️ It is given that,

x² + y2 = 34xy

➡️ Add 2xy both sides of the equation,

➡️ x² + y? + 2xy = 34xy + 2xy

➡️ (x + y )² = 36xy

➡️ (x + y )² = 6²xy

➡️ Take logarithm both sides,

➡️ log (x + y)2 = log[62xy]

➡️ 2log (x + y) = log 6²+log x + log y

➡️ [ Since 1) log a^n = n log a ,

➡️ 2) log ab = log a + log b]

➡️ 2 log (x+ y) = 2log 6 + log x + log y

➡️ 2log(x + y) = 2log(2 × 3)+logx+logy

➡️ 2 log(x + y) = 2(log2 +log3)+logx+logy

➡️ 2log(x+y)=2log2+2log3+logx+logy

➡️ Hence proved