Finding the equation of Exponential and logarithmic Functions given sufficient information.
Consider the two points (1,5) and (3,8). These points could be connected with lines represented by a
variety of functions including linear, exponential or logarithmic.
A. Find the equation of the straight line that connects two points (you’ve done this many times before).
Find the slope and then use the point-slope form of a linear equation.
=
=
y-y1=m(x-x1)
B. Finding the equation of an exponential function that connects two points can be done in two
different ways, depending on the form the function will take. Your calculator will write it in the form
y = abx. Excel will write it in the form y=aerx.
First, find the equation using the form y = abx.
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Substituted into two exponential equations 5 = ab1 and 8 = ab3.
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Divide the second equation by the first equation:
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Simplify the left side to get
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
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b = 1.2649.
Substitute b into one of the equations, 5 = ab1 so 5 = a(1.2649)1 then solve for a: a = 3.95.
The equation is y = 3.95(1.2649)x.
.
.
Second, find the equation using the form y=aerx.

Substituted into two exponential equations 5=aer1 and 8=aer3.

Divide the second equation by the first equation:

Simplify the left side:

Take the log of both sides:
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Simplify:
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Solve: r = 0.235
Substitute into one of the equations then solve for a: 5=ae(0.235)1, a = 3.95
The equation is y=3.95e0.235x .
C. Find the equation of a logarithmic Function that connects the two points. A logarithmic equation
can be written in the form y = a + b∙lnx. Then for the two points given above we can write two
logarithmic equations 5 = a + b∙ln1 and 8 = a + b∙ln3.
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
Subtract the first equation from the second: 3 = b∙ln3 - b∙ln1.
Simplify: 3 = b(ln3 – ln1)
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Since ln3 – ln1 =
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Substitute into one of the equations, 5 = a + 2.73∙ln1 then solve for a, a = 5.
The equation is y = 5 + 2.73∙lnx.
= ln 3, then
Activity 4.1 Finding exponential and logarithmic Functions Key
Name_____________________________ Effort _____/4 Attendance ___/1 Total ____/5
Complete this side in class and submit today. Complete the other side when the papers are returned for
your own practice.
Find one linear, 2 exponential and 1 logarithmic function through the points (3,2) and (7,10).
=
=
y-y1=m(x-x1)
y = abx

Substituted into two exponential equations 2 = ab3 and 10 = ab7.

Divide the second equation by the first equation:

Simplify the left side to get



b = 1.4953.
Substitute b into one of the equations, 2 = ab3 so 2 = a(1.4953)3 then solve for a: a = 0.598.
The equation is y = 0.598(1.4953)x.
.
.
y=aerx.

Substituted into two exponential equations 2=aer3 and 10=aer7.

Divide the second equation by the first equation:

Simplify the left side:

Take the log of both sides:

Simplify:



Solve: r = 0.402
Substitute into one of the equations then solve for a: 2=ae(0.402)3, a = 0.598
The equation is y=0.598e0.402x .
y = a + b∙lnx
2 = a + b∙ln3 and 10 = a + b∙ln7.


Subtract the first equation from the second: 8 = b∙ln7 - b∙ln3.
Simplify: 8 = b(ln7 – ln3)

Since ln7 – ln3 =
, then
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Substitute into one of the equations, 2 = a + 9.44∙ln3 then solve for a, a = -8.37.
The equation is y = -8.37 + 9.44∙lnx.
Find one linear, 2 exponential and 1 logarithmic function through the points (4,5) and (6,2).
=
=
y-y1=m(x-x1)
y = abx

Substituted into two exponential equations 5 = ab4 and 2 = ab6.

Divide the second equation by the first equation:

Simplify the left side to get



b = 0.632.
Substitute b into one of the equations, 2 = ab6 so 2 = a(0.632)3 then solve for a: a = 31.25.
The equation is y = 31.25(0.632)x.
.
.
y=aerx.

Substituted into two exponential equations 2=aer6 and 5=aer4.
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Divide the second equation by the first equation:

Simplify the left side:

Take the log of both sides:

Simplify:


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Solve: r = -0.458
Substitute into one of the equations then solve for a: 2=ae(-0.458)6, a = 31.25
The equation is y=31.25e-0.458x .
y = a + b∙lnx
2 = a + b∙ln6 and 5 = a + b∙ln4.


Subtract the second equation from the first: -3 = b∙ln6 - b∙ln4.
Simplify: -3 = b(ln6 – ln4)

Since ln6 – ln4 =


Substitute into one of the equations, 2 = a -7.3989∙ln6 then solve for a, a = 15.257.
The equation is y = 15.257 -7.3989∙lnx.
, then
Find one linear, 2 exponential and 1 logarithmic function through the points (2,4) and (8,9).
=
=
y-y1=m(x-x1)
y = abx

Substituted into two exponential equations 4 = ab2 and 9 = ab8.

Divide the second equation by the first equation:

Simplify the left side to get



b = 1.1447.
Substitute b into one of the equations, 4 = ab2 so 4 = a(1.1447)2 then solve for a: a = 3.053.
The equation is y = 3.053(1.1447)x.
.
.
y=aerx.

Substituted into two exponential equations 4=aer2 and 9=aer8.

Divide the second equation by the first equation:

Simplify the left side:

Take the log of both sides:

Simplify:



Solve: r = 0.135
Substitute into one of the equations then solve for a: 4=ae(0.135)2, a = 3.053
The equation is y=3.053e0.135x .
y = a + b∙lnx
4 = a + b∙ln2 and 9 = a + b∙ln8.


Subtract the first equation from the second: 5 = b∙ln8 - b∙ln2.
Simplify: 5 = b(ln8 – ln2)

Since ln8 – ln2 =


Substitute into one of the equations, 4 = a +3.6067∙ln2 then solve for a, a = 1.50.
The equation is y = 1.50 + 3.6067∙lnx.
, then