Asked by Anonymous
Show that logx+log(xy)+log(xy^2)is the arthematic progression
Answers
Answered by
Bosnian
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.
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.
Answered by
Reiny
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
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
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