3-4 Exponential and Logarithmic Equations
Solve each equation.
39. 72x + 1 = 3x + 3
SOLUTION: 41. 9x + 2 = 25x – 4
SOLUTION: 47. 25x + 6 = 42x + 1
SOLUTION: 49. ASTRONOMY The brightness of two celestial bodies as seen from Earth can be compared by determining the variation in brightness between the two bodies. The variation in brightness V can be calculated by
where mf is the magnitude of brightness of the fainter body and mb is the magnitude of brightness of
the brighter body.
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49. ASTRONOMY The brightness of two celestial bodies as seen from Earth can be compared by determining the variation in brightness between the two bodies. The variation in brightness V can be calculated by
f
b is the magnitude of brightness of
the brighter body.
3-4 Exponential where m
and Logarithmic
is the magnitudeEquations
of brightness of the fainter body and m
a. The Sun has m = −26.73, and the full Moon has m = −12.6. Determine the variation in brightness between the Sun
and the full Moon.
b. The variation in brightness between Mercury and Venus is 5.25. Venus has a magnitude of brightness of –3.7.
Determine the magnitude of brightness of Mercury.
c. Neptune has a magnitude of brightness of 7.7, and the variation in brightness of Neptune and Jupiter is 15,856.
What is the magnitude of brightness of Jupiter?
SOLUTION: a.
b.
c.
Solve each equation.
51. e2x − 15ex + 56 = 0
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SOLUTION: Page 2
3-4 Exponential and Logarithmic Equations
Solve each equation.
51. e2x − 15ex + 56 = 0
SOLUTION: 53. 6e2x − 5ex = 6
SOLUTION: The logarithm of a negative number provides no real solution, so x
0.41.
55. 8e4x − 15e2x + 7 = 0
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3-4 Exponential and Logarithmic Equations
The logarithm of a negative number provides no real solution, so x
0.41.
55. 8e4x − 15e2x + 7 = 0
SOLUTION: Solve each logarithmic equation.
61. ln x + ln (x + 7) = ln 18
SOLUTION: Substitute into original equation to eliminate extraneous solutions.
The logarithm of a negative number provides no real solution, so x cannot equal −9. The solution is 2.
63. ln (x – 3) + ln (2x + 3) = ln (−4x2)
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3-4
Substitute into original equation to eliminate extraneous solutions.
Exponential
Logarithmic
Equations
The
logarithm of aand
negative
number provides
no real solution, so x cannot equal −9. The solution is 2.
63. ln (x – 3) + ln (2x + 3) = ln (−4x2)
SOLUTION: Substitute into original equation to eliminate extraneous solutions.
The logarithm of a negative number provides no real solution, so there is no solution.
65. log (x + 6) = log (8x) − log (3x + 2)
SOLUTION: Substitute into original equation to eliminate extraneous solutions.
The logarithm of a negative number provides no real solution, so there is no solution.
67. ln (3x2 – 4) + ln (x2 + 1) = ln (2 – x2)
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Substitute into original equation to eliminate extraneous solutions.
3-4 Exponential and Logarithmic Equations
The logarithm of a negative number provides no real solution, so there is no solution.
67. ln (3x2 – 4) + ln (x2 + 1) = ln (2 – x2)
SOLUTION: Solve each logarithmic equation.
69. log2 (2x – 6) = 3 + log2 x
SOLUTION: Substitute into original equation to eliminate extraneous solutions.
The logarithm of a negative number provides no real solution, so there is no solution.
71. log x = 1 – log (x – 3)
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Substitute
original
equation to eliminate extraneous solutions.
−2 − 3 = −5
The logarithm of a negative number provides no real solution, so x = 5.
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Substitute into original equation to eliminate extraneous solutions.
3-4 The
Exponential
Logarithmic
Equations
logarithm of aand
negative
number provides
no real solution, so there is no solution.
71. log x = 1 – log (x – 3)
SOLUTION: Substitute into original equation to eliminate extraneous solutions.
−2 − 3 = −5
The logarithm of a negative number provides no real solution, so x = 5.
73. log9 9x − 2 = –log9 x
SOLUTION: Solve each logarithmic equation.
75. log (29,995x + 40,225) = 4 + log (3x + 4)
SOLUTION: 80. TECHNOLOGY A chain of retail computer stores opened 2 stores in its first year of operation. After 8 years of eSolutions Manual - Powered by Cognero
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operation, the chain consisted of 206 stores.
a. Write a continuous exponential equation to model the number of stores N as a function of year of operation t.
Round k to the nearest hundredth.
3-4 Exponential and Logarithmic Equations
80. TECHNOLOGY A chain of retail computer stores opened 2 stores in its first year of operation. After 8 years of operation, the chain consisted of 206 stores.
a. Write a continuous exponential equation to model the number of stores N as a function of year of operation t.
Round k to the nearest hundredth.
b. Use the model you found in part a to predict the number of stores in the 12th year of operation.
SOLUTION: kt
a. The general continuous exponential equation is N = N 0e . The initial number of stores N 0 is 2. At year 1, t = 1 and
N = N 0 = 2, so the equation modeling the situation is
. Use the data to solve for k.
The equation is
b.
.
81. STOCK The price per share of a coffee chain’s stock was $0.93 in a month during its first year of trading. During
its fifth year of trading, the price per share of stock was $3.52 during the same month.
a. Write a continuous exponential equation to model the price of stock P as a function of year of trading t. Round k
to the nearest ten-thousandth.
b. Use the model you found in part a to predict the price of the stock during the ninth year of trading.
SOLUTION: a. The general continuous exponential equation is P = P0ekt. The initial stock price P0 is $0.93 or 0.93. At year 1, t =
1 and P = P0 = 0.93, so the equation modeling the situation is
The equation is P = 0.93e
b.
. Use the information to solve for k.
0.3328(t − 1)
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.
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3-4 Exponential and Logarithmic Equations
81. STOCK The price per share of a coffee chain’s stock was $0.93 in a month during its first year of trading. During
its fifth year of trading, the price per share of stock was $3.52 during the same month.
a. Write a continuous exponential equation to model the price of stock P as a function of year of trading t. Round k
to the nearest ten-thousandth.
b. Use the model you found in part a to predict the price of the stock during the ninth year of trading.
SOLUTION: a. The general continuous exponential equation is P = P0ekt. The initial stock price P0 is $0.93 or 0.93. At year 1, t =
1 and P = P0 = 0.93, so the equation modeling the situation is
The equation is P = 0.93e
b.
. Use the information to solve for k.
0.3328(t − 1)
.
Solve each logarithmic equation.
83. SOLUTION: 85. 36 + 3 log3 x = 60
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3-4 Exponential and Logarithmic Equations
85. 36 + 3 log3 x = 60
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