(3,27) , (5,243) Write an exponential function y=ab^x whose graph passes through the given points.

The given two points of the exponential function are (3,27) and (5,243).
To determine the exponential function
y=ab^x
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.
27=ab^3
27/b^3=a
Plug-in this to EQ2.
243=ab^5
243=27/b^3*b^5
And, solve for b.
243= 27b^2
243/27=b^2
9=b^2
+-sqrt9=b
+-3=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.
27=ab^3
27=a(3)^3
And, solve for a.
27=27a
27/27=a
1=a
Then, plug-in a=1 and b=3 to
y=ab^x
So this becomes
y=1*3^x
y=3^x
Therefore, the exponential function that passes the given two points is y=3^x .