How do you calculate #log_3 45 - log_3 9#?

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Answer 1 · 2018-06-26T15:19:05

We will use logarithmic rules to find that the answer is #log5/log3#, or about #1.645#.

The first rule we'll use is #log_a b-log_a c=log_a (b/c)#:

#log_3 45-log_3 9=log_3 (45/9)=log_3 5#

This logarithm doesn't look too friendly. Fortunately, we can use another rule:

#log_b c=(log_a c)/(log_a b)#

It doesn't matter what base we choose for #a#, as long as each logarithm as the same base:

#log_3 5=(log_10 5)/(log_10 3)~~1.465#

Hope this helps!