Question

xsquare+y square=7xythen show that
2log(x+y)=log x+log y +2log 3​

Answer 1

Step-by-step explanation:

Given:

${x}^{2} + {y}^{2} = 7xy$

To Prove:

$2 \: log(x + y) = log(x) + log(y) + 2 \: log(3)$

Proof:

We have,

${x}^{2} + {y}^{2} = 7xy$

Adding '2xy' both sides,

$= > {x}^{2} + {y}^{2} + 2xy = 9xy$

$= > (x + y) ^{2} = 9xy$

Taking log both sides,

$= > log((x + y)^{2} ) = log(9xy)$

Using various properties of log,i.e,

$log( {x}^{n} ) = n \: log(x) \: \: and \: \: log(xy) = log(x) + log(y)$

Now, we have,

$2 \: log(x + y) = log(9) + log(x) + log(y)$

$2 \: log(x + y) = log( {3}^{2} ) + log(x) + log(y)$

$2 \: log(x + y) = 2 \: log(3) + log(x) + log(y)$

Hence, proved