Question

Evaluate log2 1 over 64.
−6

−32

6

32

Answer 1

`\log_ab=c\iff a^c=b\\\log_ab^c=c\log_ab\\\log_aa=1\\a^(-1)=(1)/(a)\\\\\log_2(1)/(64)=\log_264^(-1)=-\log_264=-\log_22^6=-6\log_22=-6`

Answer 2

The correct answer among the choices provided is the first option. When log (base 2) 1/64 is evaluated, the answer is -6. The first step is to change the logarithmic function into an exponential form. 2^(-6) = 1/64 is the correct exponential function.