How do you evaluate #log_4 128#?

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Answer 1 · 2016-12-05T02:55:41

#3.5#

Sol 1 )

Recall : #Log_a a=1, and log_a a^b=blog_aa=b#,

#128=4^3*2=4^3*4^(1/2)=4^(3+1/2)=4^(7/2)#

#=> log_4 128=log_4 4^(7/2)=(7/2)log_4 4=7/2=3.5#

Sol 2)

Let #log_4 128=x#

rewrite in exponential form :

#4^x=128#

# 4^x=4^(7/2)#

#=> x=7/2=3.5#