Question 1109516
I believe the answer is C.
{{{ f(x) = 16*(1/2)^((x-1)) }}}
Note that the {{{ x-1 }}} is the
exponent of {{{ 1/2 }}}
----------------------------
{{{ x=1 }}}
{{{ f(x)  = 16*(1/2)^((1-1)) }}}
{{{ f(x) = 16*(1/2)^0 }}}
{{{ f(x) = 16 }}}
----------------------------
{{{ x=2 }}}
{{{ f(x) = 16*(1/2)^((2-1)) }}}
{{{ f(x) = 16*(1/2) }}}
{{{ f(x) = 8 }}}
----------------------------
{{{ x=3 }}}
{{{ f(x) = 16*(1/2)^((3-1)) }}}
{{{ f(x) = 16*(1/4) }}}
{{{ f(x) = 4 }}}
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And so on
Note that this is only true if you let {{{ x }}} 
vary from 1,2,3,4, . . .