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December 18, 2014

December 18, 2014

Posted by **Kyle** on Saturday, October 27, 2012 at 4:06pm.

log_2 1/16

- calculus (math) -
**Damon**, Saturday, October 27, 2012 at 4:32pmbase^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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