How do you evaluate #log_2 (0.25)#?

Question

13256 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-22T21:04:51

#log_2(0.25)##=-2#

#log_2(0.25)#

#=log_2(1/4)#

#=log_2 (1/2^2)#

#=log_2 2^-2#->use property #x^-n=1/x^n#

#=-2log_2 2#-> use property #log_bx^n=nlog_bx#

#=-2 (1)#-> use property #log_b b=1#

#=-2#