Question

If x² + y² = 6xy, prove that 2 log (x + y) = logx + logy + 3 log 2

Answer 1

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

but x^2+y^2=6xy

so

(x+y)^2=8xy

now apply logarithm on both sides

2 log(x+y)= log(8xy)

=log8+logx+logy

=2 log3+logx+logy

Answer 2

Hey there!

Given that,

x² + y² = 6xy

Add 2xy on both sides,

x² + y² + 2xy = 6xy + 2xy

x² + y² + 2xy = xy ( 6 + 2 )

( x + y) ² = 8xy

Apply logarithm on both sides,

log ( x + y) ² = log8xy

2 log ( x + y) = log ( 2³ * xy)

2 log ( x + y) = log2³ + log xy

2 log ( x + y) = 3log2 + logx + logy

Formulae used :

logm^n = nlogm

logmn = log m + log n

a² + b² + 2ab = ( a + b)²