How do you simplify #log_20 (8000^x)#?

1 Answer
EZ as pi
Sep 11, 2016

#3x#

Explanation:

Recall:

The " power law " of logs states:

#log_p q^color(red)(m) = color(red)(m)log_p q#

The " change of base law " states

#log_a b = (log_c b)/(log_c a)#

(#"We usually use " log_10" as " log_c#)

Apply these two laws to the question:

#log_20 8000^color(red)(x) = color(red)(x)log_20 8000#

=#color(red)(x)((log_10 8000)/(log_10 20)) " "larr# use a calculator

=#x xx 3 = 3x#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
However, using another approach - the maths is beautiful, and we do not even need a calculator this time!

Note that #20^3 = 8000#

#log_20 8000^x = log_20 (20^3)^x#

=#log_20 20^(3x)#

=#3x log_20 20 " "larr log_20 20 = 1#

#3x#

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