Show that logx+log(xy)+log(xy^2)is the arthematic progression

Answer 1

log ( x ∙ y ) = log x + log y

log ( x ∙ y² ) = log x + log ( y² ) = log x + 2 log y

log x + log ( x ∙ y ) + log ( x ∙ y² ) = log x + log x + log y + log x + 2 log y

In A.P:

an = a1 + ( n - 1 ) ∙ d

in this case:

a1 = log x , d = log y

Proof:

a1 = log x

a2 = a1 + ( 2 - 1 ) ∙ d = a1 + 1 ∙ d = a1 + d = log x + log y

a3 = a1 + ( 3 - 1 ) ∙ d = a1 + 2 ∙ d = a1 + 2 d = log x + 2 log y

So log x + log ( x ∙ y ) + log ( x ∙ y² ) is arithmetic progression.

Answer 2

If a, b, and c are in an arithmetic progression, then b-a = c - b
so we have to show that log(xy) - logx = log (xy^2) - log (xy)

LS = log xy - logx
= logx + logy - logx = logy

RS = log(xy^2) - log(xy)
= logx + 2logy - logx - logy = logy
LS = RS

Yes, it is an AP