Answer:

Explanation:
meants when
, that the value for
is 5.
So this gives us this equation:

meants when
, that the value for
is 10.
So this gives us this equation:

So I take equation 2 and divide it be equation 1 I get:

Simplifying:

Since the base for an exponential function can't be negative then
.
So plugging into one of my equations I began with gives me an equation to solve for the initial value,
:


Divide both sides by 2:

The function is:
