Express log2(6!) in form a + log2(b) where a and b are integers and b is the smallest possible value?

Express #log_2(6!)# in the form #a + log_2(b)#, where a, b #in# z, and b is the smallest possible value

1 Answer
Guillaume L.
May 19, 2018

#log_2(6!)==4+log_2(45)#

Explanation:

#log_2(6!)=log_2(2*3*4*5*6)#
#=log_2(2*4)+log_2(3*5*2*3)#
#=log_2(2*4)+log_2(2)+log_2(3*5*3)#
#=log_2(2*4*2)+log_2(3*5*3)#
#=log_2(16)+log_2(45)#
#=cancel(log_2(2^4))^(color(red)(=4))+log_2(45)#
#=4+log_2(45)#

Because #gcd(45,2)=1#, #45# is the smallest #b# you can have.
\0/ here's our answer !

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