The given two points of the exponential function are (3,27) and (5,243).
To determine the exponential function
plug-in the given x and y values.
For the first point (3,27), the values of x and y are x=3 and y=27. Plugging them, the exponential function becomes:
`27=ab^3` (Let this be EQ1.)
For the second point (5,243), the values of x and y are x=5 and y=243. Plugging them, the function becomes:
`243=ab^5` (Let this be EQ2.)
To solve for the values of a and b, apply substitution method of system of equations. To do so, isolate the a in EQ1.
Plug-in this to EQ2.
And, solve for b.
Take note that in the exponential function `y=ab^x` , the b should be greater than zero `(bgt0)` . When` b lt=0` , it is no longer an exponential function.
So, consider only the positive value of b which is 3.
Now that the value of b is known, plug-in it to EQ1.
And, solve for a.
Then, plug-in a=1 and b=3 to
So this becomes
Therefore, the exponential function that passes the given two points is `y=3^x` .