Leta,b be all real #'s - {0}. Prove that log_a(x) = log_b(x)/log_b(a) by showing that a^log_b(x)/log_b(a) = x

Answer 1

By the definition of logarithm, we have log_b(x) = log_b(a^log_a(x)). Applyint the properties of the log function, we get log_b(x) = log_a(x) * log_b(a), so that log_b(x)/log_b(a) = log_a(x)

By the definition of log, we then have a^l(og_b(x)/log_b(a) = a^(log_a(x)) = x, which means that log_a(x) = log_b(x)/log_b(a)

Answer 2

First, an easy equality: b = a ^ (1/log_b(a)). Proof:

b = b ^ (log_b(a)/log_b(a))
= (b ^ (log_b(a))) ^ (1/log_b(a))
= a ^ (1/log_b(a)), since by definition of the log, b ^ (log_b(a)) = a.

So, back you your question...

a^log_b(x)/log_b(a)

= (a ^ (1/log_b(a))) ^ (log_b(x)). Notice that the stuff on the left is precisely what we have in our equality above, so this becomes...

= b ^ (log_b(x))

= x

QED.