Question 1148543
{{{f(x) = C*a^x}}}

At x = 0, f(x) = 2
{{{2 = C*a^0}}}
{{{2 = C*1}}}
{{{C = 2}}}


Therefore, {{{f(x) = 2*a^x}}}


At x=13, f(x) = 16
{{{16 = 2*a^3}}}
{{{a^3 - 8 = 0}}}
{{{(a - 2)(a^2 + 2a + 4)=0}}}


The value of {{{a}}} must be real
{{{a - 2 = 0}}} i.e. {{{a = 2}}}


So the function is
{{{f(x) = 2*2^x}}}