Question

solve the given equation

Answer 1

Given Equation is:

$log \frac{x+y}{2} = \frac{1}{2} (log x + log y)$

We know that log a + log b = log ab.

$log (\frac{x + y}{2}) = \frac{1}{2} (log xy)$

$log (\frac{x + y}{2}) = \frac{1}{2} (log xy)$

$2 log (\frac{x + y}{2} )= log(xy)$

$(\frac{x + y}{2} )^2 = xy$

$\frac{1}{4} (x + y)^2 = xy$

(x + y)^2 = 4xy

(x + y)^2 - 4xy = 0

(x - y)^2 = 0

x - y = 0

x = y.



Hope this helps!

Answer 2

Hi

Please see the attached file!

Thanks