How do you evaluate #\log _ { 2} 64- \log _ { 2} 32+ \log _ { 2} 2#?

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Answer 1 · 2016-12-27T04:35:04

#2#

Use the rules #log_an + log_am = log_a(n xx m)# and #log_an - log_am = log_a(n/m)#.

#=log_2(64/32 * 2)#

#=log_2 4#

Use the change of base rule of logarithms: #log_a n= logn/loga#.

#=log4/log2#

Write in the same base.

#=log2^2/log2^1#

Use the rule #loga^n = nloga#.

#=(2log2)/(1log2)#

#=2#

Hopefully this helps!