Question

Given that 2 log2

(x − y) − log2 x = 4 + log2 y , show that x2 + y2 = 18xy.

Answer 1

Step-by-step explanation:

Given :-

2 log2 (x-y) - log2 x = 4 + log2 y

To find :-

Show that x²+y² = 18xy .

Solution :-

Given that

2 log2 (x-y) - log2 x = 4 + log2 y

=> 2 log2 (x-y) - log2 x - log2 y = 4

=> log2 (x-y)² - log2 x - log2 y = 4

Since log a^m = m log a

=> log2 (x-y)² - (log2 x + log2 y) = 4

=> log2 (x-y)² - log2 (xy) = 4

Since log a + log b = log (ab)

=> log2 [(x-y)²/(xy)] = 4

Since log a / log b = log (a-b)

On writing it into power notation then

=> (x-y)²/(xy) = 2⁴

Since loga N = x => a^x = N

=> (x-y)²/(xy) = 2×2×2×2

=>(x-y)²/(xy) = 16

=> (x-y)² = 16×xy

=> (x-y)² = 16xy

=> x²-2xy+y² = 16xy

Since (a-b)² = a²-2ab+b²

=> x²+y² = 16xy+2xy

=> x²+y² = (16+2)xy

=> x²+y² = 18xy

Hence, Proved.

Answer:-

If 2 log2 (x-y) - log2 x = 4 + log2 y then

x²+y² = 18xy

Used formulae:-

→ log a^m = m log a

→ log a + log b = log (ab)

→ loga N = x => a^x = N

→ log a / log b = log (a-b)

→ (a-b)² = a²-2ab+b²