How do you expand #ln((2^-2)(3^1))/ln((2)(3))^-4#?

1 Answer
Tazwar Sikder
Sep 13, 2016

We have: #(ln((2^(- 2)) (3^(1)))) / (ln((2) (3))^(- 4))#

#= (ln(((1) / (4)) (3))) / (ln(((2) (3))^(- 4)))#

#= (ln((3) / (4))) / (ln(6^(- 4))#

Using the laws of logarithms:

#= (ln((3) / (4))) / (- 4 ln(6))#

#= - (ln((3) / (4))) / (4 ln(6))#

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