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`(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` .

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