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Mathematics 11 Online
OpenStudy (anonymous):

log base 2 1/64?

OpenStudy (anonymous):

\[\log_{2} 1/64\]

OpenStudy (anonymous):

i need to solve for x

OpenStudy (lalaly):

x=-6

OpenStudy (anonymous):

log_2 1/64 = log_2 1/2^6 = log_2 2^-6 = -6 So your expression evaluates to -6 I don't know where x comes into it...?

OpenStudy (lalaly):

maybe she meant log2 1/64 = x

OpenStudy (anonymous):

Maybe..

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