Chapter 12  Algebra Coach Exercises
Algebra Coach Exercise on fractional and real exponents.
Simplify each of the following exponential expressions.
To do this draganddrop each expression into the
Algebra Coach and then click the Simplify button repeatedly.
For questions 8 to 10 finish up by clicking the Add fractions button.
Required Options settings: Set the simplification mode to “exact arithmetic”.
 x^(1/5) * x^(2/5)
 x^(3/5) / x^(1/5)
 (x^(3/5))^2
 x^1.2 * x^3
 (x^1.2)^3
 (x^5)^(0.4)
 (x^(a/b))^(c/d)
 x^a * x^(c/d)
 x^(a/c) * x^(b/c)
 x^(a/b) / x^(c/d)
Algebra Coach Exercise on logarithms.
Simplify each of the following expressions involving logarithms.
To do this draganddrop each expression into the
Algebra Coach and click the Simplify button.
 log(10^(3))
 log(b^x)
 log(b^10 * c^15)
 ln(e^5)
 log(b^(3m) * c^(4m))
 ln(b^x)
 log(32)/log(8)
Write each of the following expressions as a single logarithm.
To do this draganddrop each expression into the
Algebra Coach and click the Add Logarithms button (sometimes several times).
 log(5) + log(8)
 log(5) + log(2)
 log(x) + log(y) + log(z)
 2*log(x) + 3*log(y) + z
 a*log(x)  b*log(y) + 2*log(w+z)
 log(x)/a + log(y)/b
 log(x+2) + log(x2)
Algebra Coach Exercise on the forms
for exponential growth and decay.
Each of the following questions contains an exponential growth function, shown in blue.
The exponential factor in the function is to be rewritten in the form
e^{ r t} where the exponential growth rate r
is to be determined.
To do this draganddrop the equation in red
into the Algebra Coach and use AutoSolve to solve it for r.
(Don’t forget to first set the Autosolve variable to r.)

y = 5 · 6^{ t
} solve
6 ^ t = e ^ (r t)

y = 2 · 5^{ t/3
} solve
5 ^ (t / 3) = e ^ (r t)
Each of the following questions contains an exponential decay function, shown in blue.
The exponential factor in the function is to be rewritten in the form
e^{ − r t} where the exponential decay rate r
is to be determined.
To do this draganddrop the equation in red
into the Algebra Coach and use AutoSolve to solve it for r.

y = 4 · 6^{ − t
} solve
6 ^ (t) = e ^ (r t)

y = 25 · (¼)^{ − t
} solve
(1 / 4) ^ (t) = e ^ (r t)
Algebra Coach Exercise on analyzing
exponential growth and decay.
 The graph of a certain exponential (growth) function contains the points (2, 3) and (6, 9).
Find the equation of this exponential function in the rate form,
y = a e^{ b x}, where the constants
a and b are to be determined. Here’s how:

Substitute the first point into the rate form. To do this
draganddrop the following text
into the Algebra Coach’s listbox and click the Substitute button:
x = 2, y = 3, y = a * e ^ (b x)
The result will be an equation containing a and b.
Set the Autosolve variable to a and solve this equation for a.
We will need to locate this solution later so type the words “Equation 1”
beside it or near it.

Now repeat the whole procedure with the second point:
Clear the Algebra Coach’s listbox and draganddrop the following text into it
and click the Substitute button:
x = 6, y = 9, y = a * e ^ (b x)
The result will be a second equation containing a and b.
Solve this equation for a and type the words “Equation 2” near it.

Clear the Algebra Coach’s listbox and drag both Equation 1 and Equation 2
(one after the other) into it. The listbox should now contain:
a = 3 / e ^ (2*b), a = 9 / e ^ (6*b)
Set the two expressions for a equal to each other by clicking the Substitute button.
Set the Autosolve variable to b, set the exact or floating point option to floating point
and solve this equation for b. The result will be b = 0.275.
The Algebra Coach automatically checks this solution by substituting it back into the
original equation. Since the original equation essentially reads a = a, this
means that a = 1.73, and that the equation of the exponential function is:
y = 1.73 e^{ 0.275 x}
 The graph of a certain exponential (decay) function contains the points
(5, 22.7) and (15, 9.14).
Find the equation of this exponential function in the rate form,
y = a e^{ − b x}, where the constants
a and b are to be determined. Here’s how:

Substitute the first point into the rate form. To do this
draganddrop the following text
into the Algebra Coach’s listbox and click the Substitute button:
x = 5, y = 22.7, y = a * e ^ ( b x)
The result will be an equation containing a and b.
Set the Autosolve variable to a and solve this equation for a.
We will need to locate this solution later so type the words “Equation 1”
beside it or near it.

Now repeat the whole procedure with the second point:
Clear the Algebra Coach’s listbox and draganddrop the following text into it
and click the Substitute button:
x = 15, y = 9.14, y = a * e ^ ( b x)
The result will be a second equation containing a and b.
Solve this equation for a and type the words “Equation 2” near it.

Clear the Algebra Coach’s listbox and drag both Equation 1 and Equation 2
(one after the other) into it. The listbox should now contain:
a = 22.7 * e ^ (5*b), a = 9.14 * e ^ (15*b)
Set the two expressions for a equal to each other by clicking the Substitute button.
Set the Autosolve variable to b and solve this equation for b. The result will be b = 0.091.
The Algebra Coach automatically checks this solution by substituting it back into the
original equation. Since the original equation essentially reads a = a, this
means that a = 35.8, and that the equation of the exponential function is:
y = 35.8 e^{ − 0.091 x}
Algebra Coach Exercise on finding the equation
of a straight line on a loglog or semilog graph.

Find the equation of the straight line in the graph to the right.
First, note that it must be of the form
i = a e^{ b t},
where a and b are to be determined,
and that it contains the points (t = 2, i = 4) and (t = 27, i = 1).
Draganddrop the following text into the
Algebra Coach’s listbox, click the Substitute button, and then Autosolve for a:
t = 2, i = 4, i = a e^(b t)
Now do the same for the following text:
t = 27, i = 1, i = a e^(b t)
Now drag the two equations that were solved for a into the textbox.
The textbox should now contain:
a = 4 / e^(2b), a = 1 / e^(27b)
Click the Substitute button, and then Autosolve for b:
The result will be b = −0.0555.
The Algebra Coach automatically checks this solution by substituting it back into the
original equation. Since the original equation essentially reads a = a, this
means that a = 4.47, and that the equation of the straight line is:
i = 4.47 e^{ − 0.0555 t}

Find the equation of the straight line in the graph to the right.
Follow the same steps as in the previous example. Note that the equation must be of the form
P = a e^{ b t},
where a and b are to be determined,
and that it contains the points (t = 4.5, P = 10) and (t = 6.6, P = 60).
Therefore the relevent equations are:
t = 4.5, P = 10, P = a e^(b t)
t = 6.6, P = 60, P = a e^(b t)
and the solution is:
P = 0.215 e^{ 0.853 t}

Find the equation of the straight line in the graph to the right.
Follow the same steps as in the previous example. Note that the equation must be of the form
C = a A^{ b},
where a and b are to be determined,
and that it contains the points (A = 4.5, C = 10) and (A = 6.6, C = 60).
Therefore the relevent equations are:
A = 2, C = 1, C = a A^b
A = 30, C = 9, C = a A^b
and the solution is:
C = 0.570 A^{ 0.811}

Find the equation of the straight line in the graph to the right.
Follow the same steps as in the previous example. Note that the equation must be of the form
C = a A^{ b},
where a and b are to be determined,
and that it contains the points (A = 4.5, C = 10) and (A = 6.6, C = 60).
Therefore the relevent equations are:
L = 0.1, Q = 10, Q = a L^b
L = 3, Q = 1, Q = a L^b
and the solution is:
Q = 2.10 L^{ − 0.677}
Algebra Coach Exercise on exponential equations.
Solve each of the following exponential equations.
To do this draganddrop each of them into the
Algebra Coach and click the AutoSolve button.
 9 ^ x = 27
 2.24 ^ (x + 2) = 12.5
 4.12 e ^ (3 x) = 114
 3.26 ^ x = 86.8
 e ^ (2 x) = 125
 1.05 e ^ (4 x + 1) = 5.96
 e ^ (2 x  1) = 3 e ^ (x + 3)
 5 ^ (2 x) = 8 ^ (3 x  2)
 10 ^ (4 x) = 4 * 10 ^ x
 2 ^ (5 x + 1) = 3 ^ (2 x + 1)
 7 e ^ (1.5 x) = 2 e ^ (2.4 x)
 5 ^ (2 x) = 3 ^ (3 x + 1)
 3 ^ (x ^ 2) = 175 ^ (x  1)
 e ^ x + e ^ (x) = 5
 e ^ x + e ^ (x) = 2 (e ^ x  e ^ (x))
 e ^ (4 x)  2 e ^ (2 x)  3 = 0
 e ^ (6 x)  e ^ (3 x)  2 = 0
 e ^ (6 x)  3 e ^ (3 x) + 2 = 0
Algebra Coach Exercise on logarithmic equations.
Solve each of the following logarithmic equations.
To do this draganddrop each of them into the
Algebra Coach and click the AutoSolve button.
 log(2 x + 25) = 2
 log(3 x + x ^ 2) = 1
 2log(x + 1)=3
 log(8x) = log(24)
 ln(8) + ln(x  2) = ln(3x  2)
 ln(5x + 2)  ln(x + 6) = ln(4)
 ln(x) + ln(x + 2) = 1
 2log(x)  log(1  x) = 1
 log(4x  3) + log(5) = 6
 ln(x)  2ln(x) = ln(64)
 ln(x + 2)  ln(36) = ln(x)
 log(x) + log(4x) = 2
 log(8x^2)  log(4x) = 2.54
 2log(x)  1 = log(20  2x)
 ln(2x)  ln(4) + ln(x  2) = 1