Given 1/(log(basea) x),1/(log(base b)x),1/(log(base c)x) the terms of arithmetic progression, show that a,b,c are the terms of geometric progression.

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The members `a,b,c` are in geometric progression if the following relation holds, such that:

`b^2 = a*c`

The problem provides the information that the following logarithms are the consecutive terms of an arithmetic progression, such that:

`1/(log_b x) = (1/(log_a x) + 1/(log_c x))/2`

You need to use the following trigonometric identity, such that:

`1/(log_b a) = log_a b`

Reasoning by analogy, yields:

`1/(log_a x) = log_x a`

`1/(log_b x) = log_x b`

`1/(log_c x) = log_x c`

Replacing `log_x a` for `1/(log_a x), log_x b` for `1/(log_b x), log_x c` for `1/(log_c x)` , yields:

`log_x b= (log_x a + log_x c)/2`

`2log_x b= (log_x a + log_x c)`

You need to use the power property of logarithms for the left side term, such that:

`log_x b^2 = log_x a + log_x c`

You need to convert the sum of logarithms into a logarithm of product, such that:

`log_x b^2 = log_x (a*c)`

You should notice that the bases of logarithms are equal, hence, you need to equate the arguments, such that:

`b^2 = a*c`

Hence, the last line proves that the terms `a,b,c` are in geometric progression, under the given conditions.

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