Question

If (x - y)^2 = 121xy, then 2 log (x + y) =
1) log x + log y + 3 log 2
3) log x + log y + 3 log 5
2) log x + log y + 2 log 2
4) log x + log y + 4 log 2​

Answer 1

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

Given equation is

$\rm \: {(x - y)}^{2} = 121xy$

can be rewritten as

$\rm \: {x}^{2} + {y}^{2} - 2xy = 121xy$

$\rm \: {x}^{2} + {y}^{2} = 121xy + 2xy$

$\rm \: {x}^{2} + {y}^{2} = 123xy$

On adding 2xy on both sides, we get

$\rm \: {x}^{2} + {y}^{2} + 2xy = 123xy + 2xy$

$\rm \: {(x + y)}^{2} = 125xy$

$\rm \: {(x + y)}^{2} = {5}^{3} xy$

On taking log on both sides, we get

$\rm \: log{(x + y)}^{2} = log[ {5}^{3} xy]$

We know that

$\boxed{\tt{ log {x}^{y} = ylogx \: }}$

and

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

So, using these results, we get

$\rm \: 2log(x + y) = log {5}^{3} + logx + logy$

can be rewritten as

$\rm \: 2log(x + y) = 3log5 + logx + logy$

Hence,

Option (3) is correct.

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ADDITIONAL INFORMATION

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

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

$\boxed{\tt{ \: log_{ {x}^{z} }( {x}^{y} ) = \frac{y}{z} \: }}$

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

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

$\boxed{\sf{ \: \: ln( {e}^{x}) = x \: }}$

Answer 2

Answer:

Question :-

$\mapsto$ If (x - y)² = 121xy, then 2log(x + y) is ?

Options :

● 1) logx + logy + 3log2

● 2) logx + logy + 2log2

● 3) logx + logy + 3log5

● 4) logx + logy + 4log2

Given :-

$\mapsto$ If (x - y)² = 121xy.

To Find :-

$\mapsto$ What is the value of 2log(x + y).

Solution :-

$\implies \bf (x - y)^2 =\: 121xy$

As we know that :

$\clubsuit \: \: \: \sf\bold{\pink{(x - y)^2 =\: x^2 - 2xy + y^2}}$

By putting that above identities we get,

$\implies \sf x^2 - 2xy + y^2 =\: 121xy$

By adding 4xy to both sides we get,

$\implies \sf x^2 + y^2 - 2xy + 4xy =\: 121xy + 4xy$

$\implies \sf x^2 + y^2 + 2xy =\: 125xy$

$\implies \sf x^2 + 2xy + y^2 =\: 125xy$

As we know that :

$\bigstar\: \: \: \sf\bold{\purple{(x + y)^2 =\: x^2 + 2xy + y^2}}$

By putting that above identities we get,

$\implies \sf (x + y)^2 =\: 125xy$

By taking log on both sides we get,

$\implies \sf log(x + y)^2 =\: log(125xy)$

$\implies \sf 2log(x + y) =\: log(125) + log(x) + log(y)$

$\implies \sf 2log(x + y) =\: log(5^3) + log(x) + log(y)$

$\implies \sf 2log(x + y) =\: 3log(5) + log(x) + log(y)$

$\implies \sf\bold{\red{2log(x + y) =\: log(x) + log(y) + 3log(5)}}$

$\therefore$ The value of 2log(x + y) is log(x) + log(y) + 3log(5) .

Hence, the correct options is option no (3) logx + logy + 3log5 .