Question

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

Answer 1

`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`

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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`