Question

if x2+y2=18xy then prove that 2log(x-y)=4log2+logx+logy.... I want answer​

Answer 1

Question :- if x² + y² = 18xy then prove that 2log(x - y) = 4log2 + log x+log y ?

Solution :-

→ x² + y² = 18xy

Subtracting 2xy from both sides,

→ x² + y² - 2xy = 18xy - 2xy

→ (x² + y² - 2xy) = 16xy

comparing LHS part with a² + b² - 2ab = (a - b)²,

→ (x - y)² = 2⁴xy

Taking log both sides now, we get,

→ log{(x - y)²} = log(2⁴xy)

in LHS now, using :-

• log(a^b) = b * log(a)

in RHS , using :-

• log(a * b * c) = log(a) + log(b) + log(c)
• log(a^b) = b * log(a)

we get :-

→ 2 * log(x - y) = log(2⁴) + log(x) + log(y)

Or,

→ 2log(x - y) = 4log2 + log x+log y . (Proved).

Answer 2

.

2. If

2 +

2 = 18 then show that 2 log( + ) = + +

22 + 5