Math 121.
Exponential and Logarithmic Equations
Fall 2016
Instructions. Work in groups of 3 to solve the following problems. Turn them in at the end of class
for credit.
Names.
1. Find the (a) exact solution to the exponential equation 94x−5 = 75, and then (b) approximate
the solution to 5 decimal places using a calculator.
Solution: (a) Using natural logarithms we find
ln(94x−5 = ln(75) ⇒ (4x − 5) ln(9) = ln(75) ⇒ 4x − 5 =
ln(75)
ln(9)
1 ln(75)
+5 .
4 ln(9)
(b) Therefore,
and so x =
1
x=
4
ln(75)
+5
ln(9)
≈ 1.74124
2. (a) Use the change of base formula to convert log11 36 to a logarithm in base b = 9.
(b) Convert log9 450 to the natural logarithm, and then approximate it to 5 decimal places.
Solution: (a) log11 36 =
(b) log9 450 =
log9 36
log9 11
ln 450
≈ 2.78044.
ln 9
3. (a) Find the exact solution to the equation
solution.
ex − 0.2e−x
1
= . You don’t have to simplify the exact
x
−x
e + 4e
3
(b) Use a calculator to express your answer in (a) to six decimal places.
Solution: (a) Clearing the denominators and then moving like terms to the same sides yields:
3ex − 0.6e−x = ex + 4e−x ⇒ 3ex − ex = 0.6e−x + 4e−x ⇒ (3 − 1)ex = (0.6 + 4)e−x
Multiplying both sides by ex /2 yields
4.6
(e )(e ) =
2
x
x
(b) x ≈ 0.416455.
⇒ e
2x
4.6
=
2
⇒ 2x = ln
4.6
2
1
⇒ x = ln
2
4.6
2
4. Solve the equation 42x−5 = 13−3x−4 for x. Leave answer in exact form.
Solution: Taking the natural log of both sides of the equation and distributing we obtain
(2x − 5)(ln 4) = (−3x − 4)(ln 13) ⇒ (2 ln 4)x − 5 ln 4 = (−3 ln 13)x − 4 ln(13)
now bring all the x’s to the left side, and all the numbers to the right side to obtain
(2 ln 4 + 3 ln 13)x = 5 ln 4 − 4 ln 13
and then dividing both sides by 2 ln 4 + 3 ln 13 we have
x=
5 ln 4 − 4 ln 13
2 ln 4 + 3 ln 13
5. A logarithm loga M where M > 0 can always be changed to another base b as follows.
y = loga M
⇒ M = ay
⇒ logb M = y logb a
Therefore solving this last equation for y yields
loga M =
logb M
logb a
This is called the change of base formula for logarithms.
(a) Use the change of base formula to convert log7 25 to a logarithm in base b = 11.
(b) Convert log7 200 to the natural logarithm, and then approximate it to 5 decimal places.
Solution: (a) log7 25 =
(b) log7 200 =
log11 25
log11 7
ln 200
≈ 2.72280.
ln 7
6. Find the exact solution(s) to the equation ex + 80e−x = 18. Verify that your solutions work.
80
Solution: Viewing the equation as ex + x = 18 we multiply both sides of the equation by ex to
e
obtain
80
(ex ) ex + x = 18ex ⇒ (ex )2 + 80 = 18ex
e
This is an equation of quadratic form, and the substitution u = ex turns this into a quadratic
equation:
u2 + 80 = 18u ⇒ u2 − 18u + 80 = 0 ⇒ (u − 8)(u − 10) = 0
This implies u = 8 or u = 10. Then ex = 8 implies x = ln(8) and ex = 10 implies x = ln(10). So
the solutions are x = ln(8), x = ln(10).
A check of the solutions is as follows:
x = ln(8) ⇒ eln(8) + 80e− ln(8) = eln(8) + 80eln(1/8) = 8 +
Page 2
80
= 8 + 10 = 18
8
as desired.
x = ln(10) ⇒ eln(10) + 80e− ln(10) = eln(10) + 80eln(1/10) = 10 +
80
= 10 + 8 = 18
10
as desired.
7. Solve the equation
log2 (x + 1) + log2 (x + 5) = log2 (2x + 10)
Solution: First, as long as all expressions in the logs are positive, we have
log2 [(x + 1)(x + 5)] = log2 (2x + 10)
and so (x + 1)(x + 5) = 2x + 10 and that implies
x2 + 6x + 5 = 2x + 10 and so
x2 + 4x − 5 = 0
and then
0 = x2 + 4x − 5 = (x + 5)(x − 1) = 0
and so potential solutions are x = −5 or x = 1.
All the logarithms in the original equation are defined when x = 1, so it is a solution; however
when x = −5, log(x+1) is not defined since −5+1 < 0. Thus x = −5 is not a solution. Therefore,
the only solution is x = 1.
8. Always be careful to check that your solutions work when solving logarithmic equations, because
logarithms and their properties are only defined for positive numbers. See what happens when you
solve the equation:
ln(x) + ln(x − 5) = ln(2x − 10)
For this, combine the logs on the left side using: ln(M ) + ln(N ) = ln(M N ) when M > 0 and N > 0,
and then use the property ln(E) = ln(F ) implies E = F when ln(E) and ln(F ) are defined.
Solution: First, ln(x) + ln(x − 5) = ln(2x − 10) implies ln[x(x − 5)] = ln(2x − 10) and thus
x(x − 5) = 2x − 10. Then
x2 − 5x − 2x + 10 = 0 ⇒ x2 − 7x + 10 = 0 ⇒ (x − 5)(x − 2) = 0
Therefore, x = 5 and x = 2 are “proposed solutions.” However, neither of them works because
when x = 5, ln(x − 5) = ln(0) which is not defined, and when x = 2, ln(x − 5) = ln(2 − 5) which
is undefined since 2 − 5 < 0. Thus neither of the proposed solutions work, and so there is no
solution.
Page 3
9. Solve the logarithmic equation 2 + log6 (3x − 2) = log6 (7x + 1).
Solution: We can write the equation as
log6 (7x + 1) − log6 (3x − 2) = 2 ⇒ log6
7x + 1
3x − 2
=2
converting this to exponential form the equation becomes
7x + 1
= 62
3x − 2
and so
7x + 1
= 62 ⇒ 7x + 1 = 36(3x − 2) ⇒ 1 + 72 = 108x − 7x
3x − 2
73
⇒ 73 = 101x and so x =
101
73
Therefore, x =
is the proposed solution. Because both 3x − 2 > 0 and 7x + 1 > 0 when
101
73
73
x=
, this means x =
is the solution.
101
101
10. The population of a city is currently 325000 and is expected to grow at a rate of 7.5 percent
per year for the foreseeable future. Its population is given by P (t) = 325000(1.075t ) where t is the
number of years from today.
(a) What will the population be in 9 years? (Express answer as a whole number)
(b) At this rate of growth, how long (in years) will it take the population to double? How long (in
years) would it take the population to quadruple? Express answers to 1 decimal place.
(c) If this growth rate could continue, how long (in years) would it take for the population to reach
3,000,000 people? Express answer to 1 decimal place.
Solution: (a) The population will be
P (9) = 325000(1.0759 ) ≈ 623103 in 9 years.
(b) To determine how long it will take the population to double, we solve 2(325000) =
325000(1.075)t for t. Thus dividing both sides by 325000 we have 2 = 1.075t and then
log(2) = t log(1.075)
⇒
t=
log(2)
≈ 9.58436
log(1.075)
Thus it would take approximately 9.6 years for the population to double.
For the population to quadruple, it would have to double twice, so it would take 2(9.58436) ≈ 19.2
years for the population to quadruple.
(c) To determine how long it will take the population to reach 3,000,000, we solve 3000000 =
325000(1.075)t for t. Thus dividing both sides by 325000 we have 3000000/325000 = 1.075t and
then
log(3000000/325000)
log(3000000/325000) = t log(1.075) ⇒ t =
≈ 30.73178
log(1.075)
Thus it would take approximately 30.7 years for the population to reach 3,000,000 people.
Page 4
10. The base can always be changed in an exponential function using the property a = blogb a ; and
therefore ax = (blogb a )x = bx logb a .
(a) Write the function f (x) = 13x in as an exponential function in base 9.
(b) Write the function g(x) = 22(13x ) using the natural base e.
Solution: (a) f (x) = 13x = 9x log9 13 .
(b) g(x) = 22ex ln 13 .
Page 5