Solve. It may be helpful to write the equation into exponential form and use the like bases property.
log
2
(
1
128
)
=
m
\displaystyle {{\log}_{{2}}{\left(\frac{{1}}{{128}}\right)}}={m}
lo
g
2
(
128
1
)
=
m
m
=
\displaystyle {m}=
m
=
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