How do you approximate #log_3 (2/3)# given #log_3 2=0.6310# and #log_3 7=1.7712#?

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Answer 1 · 2016-11-01T13:05:07

#log_3(2/3) = -0.369#

Using the difference rule of logarithms, #log_a(n/m) = log_an - log_am#, we can rewrite #log_3(2/3)# as #log_3(2) - log_3(3)#

We know that #log_3 2 = 0.6310#.

#=>0.6310 - log_3 3#

Let's use the change of base rule to see what #log_3 3# can be simplified to.

#log_3(3) = log3/log3 = 1#

So, #log_3(2/3) = 0.6310 - 1 = -0.369#

Hopefully this helps!