# Evaluate `( log_a b)( log_b c)( log_c a)`

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### 2 Answers

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.`

To evaluate `(log_ab)(log_bc)(log_ca)`, we have to use the change of base formula of logarithms such that :

`log_ax=(log_bx)/(log_ba)` where, it is assumed that x, a and b are positive real numbers so that `a!=1` , `b!=1` (also x>0).

So, converting to the base e, we can write the given expression as:

`(log_ab)(log_bc)(log_ca)=((log_eb)/(log_ea))((log_ec)/(log_eb))((log_ea)/(log_ec))`

`=(lnb/lna)*(lnc/lnb)*(lna/lnc)`

`=1`

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