How do you approximate #log_5 (2/3)# given #log_5 2=0.4307# and #log_5 3=0.6826#?

1 Answer
Oct 28, 2016

Answer:

#log_5(2/3) ~= -0.2519#

Explanation:

We can write #log_5(2/3)# as #log_5(2) - log_5(3)# using the rule quotient rule of logarithms that #log_a(m/n) = log_a(m) - log_a(n)#.

#log_5(2/3) = log_5(2) - log_5(3) = 0.4307 - 0.6826 = -0.2519#

Hopefully this helps!