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 :

$4^x=128$

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

$=> x=7/2=3.5$

How do you evaluate log_4 128log_4 128?