Question

If x2+y2= 6xy
Then prove that 2log(x+y)= logx+logy+3log2

Answer 1

$\large\underline{\sf{Given- }}$

$\rm \: {x}^{2} + {y}^{2} = 6xy$

$\large\underline{\sf{To\:prove- }}$

$\rm \: 2log(x + y) = logx + logy + 3log2$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: {x}^{2} + {y}^{2} = 6xy$

On adding 2xy on both sides, we get

$\rm \: {x}^{2} + {y}^{2} + 2xy= 6xy + 2xy$

$\rm \: {(x + y)}^{2} = 8xy$

Taking log on both sides, we get

$\rm \: log{(x + y)}^{2} = log(8xy )$

We know,

$\boxed{\sf{ \: \: log {x}^{y} \: = \: y \: logx \: }}$

$\boxed{\sf{ \: \: logxy = logx + logy \: }}$

So, using these results, we get

$\rm \: 2log(x + y) = logx + logy + log8$

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

$\rm \: 2log(x + y) = logx + logy + 3log2$

Hence, Proved

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Additional Information :-

$\boxed{\sf{ \:log \frac{x}{y} = logx - logy \: }}$

$\boxed{\sf{ \: log_{x}(y) = \frac{logy}{logx} \: }}$

$\boxed{\sf{ \: log_{x}(x) = 1 \: }}$

$\boxed{\sf{ \: log_{ {x}^{a} }( {x}^{b} ) = \frac{b}{a} \: }}$

$\boxed{\sf{ \: log_{ {x}^{a} }( {y}^{b} ) = \frac{b}{a} log_{x}(y) \: }}$

$\boxed{\sf{ \: {a}^{ log_{a}(x) } = x \: }}$

$\boxed{\sf{ \: {a}^{y log_{a}(x) } = {x}^{y} \: }}$

Answer 2

ANSWER IS IN THE ATTACHMENT ABOVE

HENCE PROVED!!!!