How do you calculate #log _2 (1/8)#?

1 Answer
Jul 27, 2016

Answer:

-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#