How do you calculate #log _2 (1/8)#?
1 Answer
Jul 27, 2016
-3
Explanation:
Let
# log_2(1/8)=n# and we want to calculate the value of n.
#color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(log_b x=nhArrx=b^n)color(white)(a/a)|)))# here b = 2 and x
#=1/8rArr1/8=2^nrArrn=-3# Since
#2^-3=1/2^3=1/8#
#rArrlog_2(1/8)=-3#
