The given
two points of the exponential function are (3,27) and (5,243).
To determine
the exponential function
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:
(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:
(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.
Plug-in this to EQ2.
And, solve for
b.
Take note that in the exponential function , the b should be greater than zero
. When
, 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.
And, solve for a.
Then, plug-in a=1 and b=3 to
So this becomes
Therefore, the exponential function that
passes the given two points is .
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