Question

plzzzx ans ....Thnk you .....Logarithms

Answer 1

Hey there !!!!

log(x+y)/2=(logx+logy)/2

log(x+y)/2=(logxy)/2

xloga=logaˣ

So,

log(x+y)/2=(log√xy)


If loga=logb then a=b

So,

log(x+y)/2=(log√xy)

(x+y)/2=√xy-------------Equation 1

Squaring equation 1

(x+y)²/4=(√xy)²

x²+y²+2xy=4xy

x²+y²+2xy-4xy=0

x²-2xy+y²=0

(x-y)²=0

x=y

Hope this helped you..............

Answer 2

log (x+y)/2 = ½(log x+log y)

2 log (x+y)/2 = log x + log y

log {(x+y)/2}² = log xy

Remove logarithms on both sides,

{(x+y)/2}² = xy

(x+y)²/4 = xy

(x+y)² = 4xy

x²+y²+2xy = 4xy

x²+y² = 4xy - 2xy

x²+y² = 2xy

x²+y²-2xy = 0

(x-y)² = 0

x-y = 0

x = y

Hence proved.

Hope it helps...