You also may solve the problem using the formula that interchanges the base with argument of logarithm, such that:

`log_a b = 1/(log_b a)`

Replacing `1/(log_b a)` for `log_a b` yields:

`1/(log_b a)*(log_b c)*(log_c a)`

You need to notice that `(log_b c)/(log_b a) = log_a c` , such that:

`(log_a...

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You also may solve the problem using the formula that interchanges the base with argument of logarithm, such that:

`log_a b = 1/(log_b a)`

Replacing `1/(log_b a)` for `log_a b` yields:

`1/(log_b a)*(log_b c)*(log_c a)`

You need to notice that `(log_b c)/(log_b a) = log_a c` , such that:

`(log_a c)*(log_c a) =(log_a c)*(1/log_a c)` (`log_c a` is replaced with `log_c a` , using the formula that interchanges the base with argument)

Reducing duplicate factors yields:

`(log_a c)*(1/log_a c) = 1`

**Hence, evlauating the product of logarithms, yields **`(log_a b)*(log_b c)*(log_c a) = 1.`