# calculus (math)

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Simplify exactly

log_2 1/16

• calculus (math) -

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

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