A strange equality

[Paolo902]

Hi guys,

I've come back to English Corner with a very simple exercise. Enjoy it.

Theorem.
$log_3 2 * log_4 3 * ... * log_8 7 = 1/3$.

Would you like to prove it?
Bye,
Paolo

[pat871]

Using the change of basis formula

$log_a b = (log_k b)/(log_k a)$

we find

$log_3 2 . ... . log_8 7 = (log_8 2)/(log_8 3). ... . (log_8 6)/(log_8 7). log_8 7 = log_8 2 =1/3$, since $8^(1/3) = 2$.