Help! (1 Viewer)

[DarkDude]

How do you prove that the logarithms of a set of numbers in G.P are in A.P?

i appreciate any help i can get on this.

 

[Jago]

T_k+1 - T_k = T_k - T_k-1

prove that and you've proved that the numbers are in AP

 

[klaw]

prove that there is a common difference (which is what jago said but I reckon mine's easier to understand)

 

[DigitalFortress]

i don't think his question is asking that, its asking about the LOGS of numbers in a GP, are in an AP.

 

[haboozin]

i don't think his question is asking that, its asking about the LOGS of numbers in a GP, are in an AP.

1 + x + x^2 + x^3... + x^n


ln1 + lnx + lnx^2 + lnx^3 + ...+ lnx^n

lnx + 2lnx + 3lnx + ... + n lnx


now as you see with Jago's method

3lnx - 2lnx = lnx

2lnx - lnx = lnx

therefore its in AP


So you see he was right... just because the question is in a certain way doesnt mean the definition of GP and AP changes.

 

[acmilan]

or more generally,

a₁ + a₂ + a₃ + ... + a_n-1 + a_n + _n+1 + ...

Since its in geometric progression, a_n/a_n-1 = a_n+1/a_n

Taking log of both sides:

ln(a_n/a_n-1) = ln(a_n+1/a_n)
ln(a_n) - ln(a_n-1) = ln(a_n+1) - ln(a_n)

which proves arithmetic progression of the logs of terms in a geometric progression

 

[DarkDude]

or more generally,

a₁ + a₂ + a₃ + ... + a_n-1 + a_n + _n+1 + ...

Since its in geometric progression, a_n/a_n-1 = a_n+1/a_n

Taking log of both sides:

ln(a_n/a_n-1) = ln(a_n+1/a_n)
ln(a_n) - ln(a_n-1) = ln(a_n+1) - ln(a_n)

which proves arithmetic progression of the logs of terms in a geometric progression

Thanks you all very much for that answer. i didnt think it was that easy.