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

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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...

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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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