(1,40) , (3, 640) Write an exponential function y=ab^x whose graph passes through the given points.

The given two points of the exponential function are (1,40) and (3,640).
To determine the exponential function 
y=ab^x
plug-in the given x and y values.
For the first point (1,40), plug-in x=1 and y=40.
40=ab^1
40=ab        (Let this be EQ1.)
For the second point (3,640), plug-in x=3 and y=640.
640=ab^3      (Let this be EQ2.)
To solve for the values of a and b, apply the substitution method of system of equations. To do so, isolate the a in EQ1.
40=ab
40/b=a
Plug-in this to EQ2.
640=ab^3
640=(40/b)b^3
And, solve for b.
640=40b^2
640/40=b^2
16=b^2
+-sqrt16=b
+-4=b
Take note that in exponential function y=ab^x , the b should be greater than zero (bgt0) . When blt=0 , it is no longer an exponential function.
So, consider on the positive value of b which is 4.
Now that the value of b is known, plug-in it to EQ1.
40=ab
40=a(4)
And, solve for a.
40/4=a
10=a
Then, plug-in the values of a and b to the exponential function
y=ab^x
So this becomes:
y= 10*4^x
Therefore, the exponential function that passes the given two points is y=10*4^x .