Question

x2+y2=27, prove that 2log (x-y) =5log2+logx+logy​

Answer 1

Step-by-step explanation:

Given Question :-

x²+y² = 27, Prove that 2 log (x-y) = 5log2 + log x + log y

Correction :-

x²+y² = 27xy

Solution :-

Given equation is x²+y² = 27xy

On subtracting 2xy both sides then

=> x²+y² -2xy = 27xy -2xy

=> x²-2xy+y² = (27-2)xy

=> (x-y)² = 25xy

Since (a-b)² = a²-2ab+b²

Where , a = x and b = y

On taking Logarithms both sides

=> log (x-y)² = log 25xy

We know that

log a^m = m log a

=> 2 log (x-y) = log 25xy

=> 2 log (x-y) = log(5²×x×y)

We know that

log ab = log a + log b

=> 2 log (x-y) = log 5² + log x + log y

We know that

log a^m = m log a

=> 2 log (x-y) = 2 log 5 + log x + log y

Hence, Proved.

Answer:-

If x²+y² = 27xy then 2 log (x-y) = 2 log 5 + log x + log y

Used formulae:-

• log ab = log a + log b

• log a^m = m log a

• (a-b)² = a²-2ab+b²

Answer 2

Given :-

$\:$

$\bf⟼ \: \: {x}^{2} + {y}^{2} \: = \: 27$

$\:$

To Prove :-

$\:$

$\bf⟼ \: \: 2 \: log \: (x-y) \: = \: 5 \: log \: 2 \: + \: log \: x \: + \: log \: y$

$\:$

Solution :-

$\:$

Given that,

$\:$

$\bf⟼ \: \: {x}^{2} + {y}^{2} \: = \: 27$

$\:$

$\bf⟼ \: \: {x}^{2} + {y}^{2} \: = \: 27xy$

$\:$

Subtracting 2xy on both sides,

$\:$

$\bf⟼ \: \: {x}^{2} + {y}^{2} - 2xy\: = \: 27 xy- 2xy$

$\:$

$\bf⟼ \: \: ({x} - {y})^{2} \: = \: 25xy \: \: \: \: \: \: \: \: \: \bigg( ∴( {a - b)}^{2} \: = \: {a}^{2} + {b}^{2} - 2ab \bigg)$

$\:$

Adding log on both sides,

$\:$

$\bf⟼ \: \: log \: ({x} - {y})^{2} \: = \: log \: 25xy$

$\:$

$\bf⟼ \: \: 2 \: log \: ({x} - {y}) \: = \: log \: {5}^{2} \: + \: log \: x \: + \: log \: y$

$\:$

$\bf⟼ \: \: 2 \: log \: ({x} - {y}) \: = \: 2 \: log \: {5} \: + \: log \: x \: + \: log \: y$

$\:$

Hence, proved !!!

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