Download CH6 Section 6.4 - Faculty Website Index Valencia College

Chapter 6
Exponential and Logarithmic Functions
and Applications
Section 6.4
Section 6.4
Exponential and Logarithmic Equations
• Solving Exponential Equations
o Exponential Equality Property
o Base 10 or Base e
o Graphically
• Fundamental Properties of Logarithms
o Solving Exponential Equations Using Properties of Logarithms
• Solving Logarithmic Equations
• Solving Literal Equations Involving Exponential and Logarithmic Equations
• Change of Base Formula
• Applications
Exponential Equality Property:
For b > 0 and b  1, if bm = bn, then m = n.
*Solving Exponential Equations with the Exponential
Equality Property:
1. Express each side of the equation as a power
of the same base.
2. Apply the exponential equality property to equate
the exponents.
3. Solve for the variable.
4. Check your solution.
*Also referred to as solving by "equating the bases" or
"relating the bases."
Use the exponential equality property to solve 292x + 4 = 162.
Step 1. Express each side of the equation as a power of the
same base.
First, we divide both sides of 292x + 4 = 162 by “2.”
92x + 4 = 81
Next, we express each side of the equation as a power of “3.”
(32)2x + 4 = 34
34x + 8 = 34
Make sure to distribute correctly: 2(2x + 4) = 4x + 8
Steps 2-3. Apply the exponential equality property to equate the
exponents. Solve for the variable.
4x + 8 = 4
x = –1
Step 4. Check your solution.
29[2(–1) + 4] = 162  292 = 162

2(81) = 192
True
x 1
Use the exponential equality property to solve 5  25 .
 
x 1
1/2
2
5
 5
51/2  52x 2
1/2  2x  2
5/4  x
Check:
5  25[(5/4) 1]
5  251/4
2.2361  2.2361
Solving Exponential Equations with Base 10 or Base e.
1. Isolate the power (the term containing the variable
exponent) on one side of the equation.
If necessary, divide both sides of the equation by
any coefficient of the power term.
2. Convert the equation to logarithmic form.
3. Solve for the variable. Use a calculator if necessary.
4. Check your solution.
Solve the exponential equation 2(–6 + 102x) = 16.26.
Round your answer to 4 decimal places as needed.
Step 1. Isolate the power.
First, divide both sides of the equation by “2.” Then, add “6” to
both sides of the equation.
–6 + 102x = 8.13
102x = 14.13
Step 2. Convert the equation to logarithmic form.
log1014.13 = 2x or log 14.13 = 2x
Step 3. Solve for the variable.
x = log 14.13
2
x  0.5751
Step 4. Check your solution.
2[–6 + 10(2 0.5751)] = 16.26  2(8.1319) = 16.2638  16.26
Solve the exponential equation 25.6 = –2 + 3.1e–0.15x. Round your
answer to 4 decimal places as needed.
27.6 = 3.1e–0.15x
27.6 = 3.1e–0.15x
3.1
3.1
8.9032 = e–0.15x
loge 8.9032 = –0.15x
or
ln 8.9032 = –0.15x
x = ln 8.9032
–0.15
x  –14.5761
You can check the answer.
Solving Exponential Equations Graphically
Use your graphing calculator to solve 2x + 5 = 32x + 6 – 1
graphically.
Can use intersection method or x-intercept method.
Let Y1 = 2x + 5 and Y2 = 32x + 6 – 1. We will use the window
[–6, 6, 1] by [– 6, 12, 1] and find the intersection.
The answer to the given equation is x = –2.
Checking: 2(–2 + 5) = 3[2(–2) + 6] – 1
8 = 8
Recall the Basic Properties of Logarithms: For b > 0 and b  1,
log x
logb 1 = 0, logb b = 1, logb bx = x, and b b  x.
Fundamental Properties of Logarithms
For positive real numbers M, N, and b, b  1, and any real
number k:
Product Property: logb (MN)  logbM  logbN
M

Quotient Property: logb    logbM  logbN
N
Power Property:
logbMk  k logbM
Note: These properties apply to natural logarithms as well.
Use the fact that log3 4 = 1.2619 and log3 7 = 1.7712 and the
properties of logarithms to estimate the value of the following
expressions. Round your answers to 4 decimal places as needed.
a. log3 28
= log3 (47) = log3(4) + log3(7) = 1.2619 + 1.7712 = 3.0331
b. log3 (1.75)
= log3 (7/4) = log3(7) – log3(4) = 1.7712 – 1.2619 = 0.5093
c. log3 4
 
 log3 41/ 2  1/2 log3 4  (1/2)(1.2619)  0.6309
Use the properties of logarithms to rewrite each expression as a
single logarithm.
a. 5 log x + log 9
= log x5 + log 9 = log (9x5)
b. log (x + 3) – log (x2 – 9)
x 3 
x 3
1 




 log  2
  log 
  log 

 x  3
 x  9
 (x  3)(x  3) 
c. 1 (2 logb x  logby)  logb 8
3
2

1
1
x
 (logb x 2  logb y)  logb 8  logb    logb 8
3
3
 y 
2 1 / 3
x
 logb  
 y 
  x 2 1/3 
 logb 8  logb 8   or
  y  


  x 2 
logb 8 3   
  y  
Expand the given expression in terms of simpler logarithms.
Assume that all variable expressions are positive real numbers.
2
log x
y
 x 2 1 / 2
 log  
 y 
 
 x2 
1
 log  
2
 y 
1
 (log x 2  log y)
2
1
 (2 log x  log y)
2
1
 log x  log y
2
Solving Exponential Equations Using Properties of
Logarithms
1. Isolate the power on one side of the equation.
2. Take the logarithm of both sides of the equation; may
take either the common logarithm (base 10) or a
natural logarithm (base e).
3. Apply the power property of logarithms to simplify
(that is, "bring down" the variable exponent to the
front).
4. Solve for the variable.
5. Check your solution.
Solve the exponential equation –7 + 4x = 5. Round your answer
to 4 decimal places.
Step 1: Isolate the power on one side of the equation.
4x = 12
Step 2: Take the logarithm of both sides of the equation.
log 4x = log 12
Step 3: Apply the power property of logarithms.
x log 4 = log 12
Step 4: Solve for the variable.
x
log 12
 1.7925
log 4
5. Check your solution.
–7 + 4(1.7925)  5
Solve the exponential equation 27ex – 2 = 7. Round your answer
to 4 decimal places.
ex – 2 = 0.2593
ln ex – 2 = ln 0.2593
(x – 2) ln e = ln 0.2593
(x – 2) (1) = ln 0.2593
(Recall ln e is equivalent to loge e)
x = ln 0.2593 + 2
x  0.6502
Checking:
27e(0.6502 – 2)  7
Logarithmic Equality Property:
For positive real numbers m, n, and b, b  1,
if logb m = logb n, then m = n.
Solving Logarithmic Equations:
1. Isolate the logarithmic expression on one side of
the equation. If needed, apply the properties of
logarithms to combine all logarithms as a single
logarithm.
2. Convert the logarithmic equation to an exponential
equation.
3. Solve for the variable.
4. Check for possible extraneous solutions.
Solve the logarithmic equation 6 log 4x = 18.
Step1. Isolate the logarithmic expression on one side of the
equation.
log 4x = 3
Step 2: Convert the logarithmic equation to an exponential
equation.
103 = 4x
Step 3: Solve for the variable.
4x = 1000
x = 250
Step 4: Check for possible extraneous solutions.
6 log [4(250)] = 18
18 = 18
20
Solve the logarithmic equation log  x    2  log (3x).
3

20
log  x    log(3x)  2
3

20
log (3x) x    2

3 

log(3x 2  20x)  2
102  3x 2  20x
3x 2  20x  100  0
(3x  10)(x  10)  0
x  10/3 or x  10
We discard x = –10/3 (it is not in the domain). Thus, x = 10.
You can verify the solution.
Solve the equation ln x + ln (x + 3) = ln (x + 15).
ln [x(x + 3)] = ln (x + 15)
ln (x2 + 3x) = ln (x + 15)
x2 + 3x = x + 15
 Logarithmic equality
x2 + 3x – x – 15 = 0
x2 + 2x – 15 = 0
(x + 5)(x – 3) = 0
x = –5 or x = 3
We discard x = –5 (it is not in the domain). Thus, x = 3.
You can verify the solution.
Solve the equation ln (2x – 1) = 3 and approximate your
answer to 4 decimal places.
Recall ln (2x – 1) = 3 is equivalent to loge(2x – 1) = 3
e3 = 2x – 1
2x = e3 + 1
2x = 21.085537
x  10.5428
Solving Literal Equations Involving Exponential or
Logarithmic Equations
Solve for y: logb x  mn  logb y
logb x  logb y  mn
x
logb    mn
y
mn
b
x

y
x
(y)(bmn )   (y)
y
y
x
bmn
Change of Base Formula
For positive real numbers M, a, and b, a  1, b  1,
logbM 
logaM
loga b
We may convert any given base into either base 10 or base e.
Example: Use the change of base formula to evaluate log2 5, and
graph y = log2 x by applying the change of base formula.
log2 5 
log 5
 2.3219
log 2
or
log2 5 
Graph of y = log2 x:
ln 5
 2.3219
ln 2
The function P(t) = 89.371(3.2)t models the number of digital 3D
screens worldwide for t number of years after 2005. Using this
model, estimate when the number of digital 3D screens worldwide
reached approximately 9,000. Solve algebraically and round your
answer to the nearest whole number. Source: mpaa.org.
89.371(3.2)t  9000
3.2t  100.7038
log 3.2t  log 100.7038
t log 3.2  log 100.7038
t
log 100.7038
4
log 3.2
The number of digital 3D screens reached approximately 9,000 in
2009.
The number of prohibited firearms intercepted by the
Transportation Security Administration (TSA) at U.S. airport
screenings from 2005 to 2009 can be modeled by the function
f(x) = 2391.893 – 931.691 ln x, where x = 1 represents 2005,
x = 2 is 2006, and so on. Use this model to estimate when the
TSA intercepted approximately 892 firearms. Round your answer
to the nearest whole number. Sources: www.aaa.com; www.safecarguide.com
2391.893  931.691 ln x  892
 931.691 ln x  1499.893
ln x  1.6099
x  e1.6099
x 5
The TSA intercepted approximately 892 firearms in 2009.
Using your textbook, practice the
problems assigned by your instructor to
review the concepts from Section 6.4.