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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