How do you write #1/3ln27 - ln(2^4-8)# as a single logarithm?

1 Answer
Jim G.
Apr 5, 2016

#ln(3/8)#

Explanation:

Using the #color(blue)" laws of logarithms " #

#• logx + logy hArr logxy #

#• logx - logy hArr log(x/y) #

#• logx^n hArr nlogx #

#rArr 1/3ln27 - ln(16 - 8 ) = ln27^(1/3) - ln8 = ln3 - ln8 #

and # ln3 - ln8 = ln(3/8) #

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