Question 920255
{{{f=b^x}}};
Your one and only data point may be insufficient.  Base two might be the choice for b.


{{{f=a*2^x}}}, including  a possible factor of "a".
{{{log(2,f)=log(2,a)+log(2,2^x)}}}
{{{log(2,f)=x*log(2,2)+log(2,a)}}}
{{{log(2,(2^-4))=2*1+log(2,a)}}}, when the one data point is substituted.
{{{-4=2+log(2,a)}}}
{{{highlight_green(-6=log(2,a))}}}


Convert that into exponential form, and it means {{{2^(-6)=a}}};
{{{highlight_green(a=1/64)}}}.


{{{highlight(f(x)=(1/64)(2^x))}}}