Prove that log a, log ar, log ar^2 is an a.p.

Is the following below correct?
Log ar^2 - Log ar= Log ar - Log a

hence applying laws of logarithm
Log(ar^2/ar) = log(ar/a)

Log and log cancels out and then cross-multiply
hence a^2r^2 = a^2r^2
L.H.S=R.H.S

Hence proved?

log a, log ar, log ar2, ...log a (r)^n

now, log ar^n=log a + n log r
consider the next term
log ar^(n+1)= log a + (n+1)log r

then

log ar^(n+1)-log ar^n= log r which is a constant. Therefor, the progression is arithemetic.

To prove that log a, log ar, and log ar^2 form an arithmetic progression (AP), we need to show that the difference between any two consecutive terms is constant.

Let's calculate the difference between the second term and the first term:

log(ar) - log(a)

Using the logarithmic property of division, we can rewrite this as:

log(ar/a)

Next, let's calculate the difference between the third term and the second term:

log(ar^2) - log(ar)

Again, applying the logarithmic property of division:

log(ar^2/ar)

Now, comparing the two expressions, we can see that they are equal:

log(ar/a) = log(ar^2/ar)

Since the logarithmic functions are equal, the terms log a, log ar, and log ar^2 satisfy the definition of an arithmetic progression (AP). The difference between any two consecutive terms is constant.

Therefore, the given sequence log a, log ar, log ar^2 is indeed an arithmetic progression (AP).

To prove that log a, log ar, log ar^2 is an arithmetic progression (A.P.), we need to show that the difference between consecutive terms is constant.

Let's start with the given logarithmic terms: log a, log ar, log ar^2.

We can express each term using the laws of logarithms:
log a = log(ar^0) (using the property log a^n = n * log a)
log ar = log(ar^1) (using the property log ab = log a + log b)
log ar^2 = log(ar^1 * r) (using the property log a^n = n * log a)

Now, let's calculate the difference between consecutive terms:
log ar - log a = log(ar^1) - log(ar^0)
= log(ar/a^0)
= log(ar)
log ar^2 - log ar = log(ar^1 * r) - log(ar^1)
= log(ar^1 * r/ar^1)
= log(r)

Therefore, the difference between consecutive terms in the sequence log a, log ar, log ar^2 is log(r).

If we compare this with the difference between the first two terms and the last two terms of the equation you provided (log ar^2 - log ar = log ar - log a), we can see that the equation is incorrect.

To prove that log a, log ar, log ar^2 is an A.P., we need to show that the difference between consecutive terms is constant, which is log(r) in this case.

Hence, the given sequence log a, log ar, log ar^2 is not an A.P.