Question

ANYBODY SCHOLAR IN MATHS ?

If x^2 + y^2 =
7xy,
then log 1/3 (x+y)=

then x is

(a) (logx+logy)

(b) ½(logx+logy)

(c) 1/3(logx+logy)

(d) 3(logx/logy)​​

Answer 1

Answer:

1/2 log(x+y), option b is correct

Given:

x² + y² = 7xy

To find:

log 1/3 (x+y) = ?

Solution:

x² + y² = 7xy

adding 2xy on both sides,

x² + y² +2xy = 7xy + 2xy

(x+y)² = 9xy

x + y = 3$\sqrt{xy}$

In place of x+y, we can put 3$\sqrt{xy}$

log(x+y /3)  =  log(3$\sqrt{xy}$/3)

                   = log$\sqrt{xy}$

                   = log$(xy)^{1/2}$

                   = 1/2 log(xy)

                   = 1/2 log(x+y)

Hence, 1/2 log(x+y) is the answer.

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