Question

Log x+y/3 =1/2log x+log) then prove x/y+y/x

Answer 1

Step-by-step explanation:

log [ ( x + y )/3] = 1/2 ( logx + logy)

2× log[ ( x + y )/3 ] = log x + log y

log [ ( x + y ) / 3 ]^2 = log xy

we know that n log a = log a^n

log a + log b = log ab }

Remove log bothsides,

[ ( x + y ) / 3 ]^2 = xy

( x + y )^2 / 3^2 = xy

x^2 + y^2 + 2xy = 9xy

x^2 + y^2 = 9xy - 2xy

x^2 + y^2 = 7xy

Divide each term with xy

x^2 / xy + y^2 / xy = 7xy / xy

x / y + y / x = 7

Answer 2

Answer:

so eto side do

Step-by-step explanation:

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