Posted by Kyle on .
Simplify exactly
log_2 1/16

calculus (math) 
Damon,
base^log x = x
2^z = 1/16
where z is log_2 (1/16)
but we know that
2^0 = 1
2^1 = 1/2
2^2 = 1/4
2^3 = 1/8
2^4 = 1/16 ah ha
so 2^(4) = 1/16
z = 4
and log_2 (1/16 ) = 4
check:
b^log_b(x) = x
2^4 = 1/16 ???
1/2^4 = 1/16 sure enough