How do you calculate #Log_2 64#?

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Answer 1 · 2018-05-02T05:54:31

#" "#
#color(blue)(log_2(64)=6#

#" "#
#color(red)(log_2(64)#.

Rewrite #color(green)64# as #color(green)(2^6#.

Use the Log Rule: #color(brown)(log_a(x^b)=[b*log_a(x)]#

#log_2(2^6) = 6*[log_2(2)]#

Since, #[log_2(2)]=1#,

#log_2(64)=6.1=6#

Hence,

#color(blue)(log_2(64)=6#

Hope it helps.