Solution Key to Assignment 14:College Algebra
These problems are from a different books and the page numbers do not correspond
to the ones of your book.
P. 402
Solve to 3 significant digits
2. e x = 9.62 ⇒ x = ln 9.62 ≈ 2.26
8.
(log 5000) + 1 (log 5) + 3 + 1 4 + log 5
=
=
≈ 1.17
4
4
4
4
12. 4 − x = 0.0001 ⇒ − x log 4 = log 0.0001 = −3 ⇒ − x log 4 = −4 ⇒ x =
≈ 6.64
log 4
8
8
8
16. log 8 − log x = 2 ⇒ log = 2 ⇒ = 10 2 = 100 ⇒ x =
= 0.08
x
x
100
10 4 x −1 = 5000 ⇒ 4 x − 1 = log 5000 ⇒ x =
ln 1.25
≈ 0.00893
25
1000
10 1000
= 1000 ⇒ −0.4 x = ln
⇒ x = − ln
≈ −7.70
46
4
46
22. e 25 x = 1.25 ⇒ 25 x = ln 1.25 ⇒ x =
24. 46e −0.4 x
Solve to get the exact value.
30.
3x + 1
3x + 1
⇒ x +1 =
⇒ x( x + 1) = 3x + 1 ⇒ x 2 − 2 x − 1 = 0
x
x
⇒ x = 1 ± 2 ; Note that x ≠ 1 − 2 , since for x = 1 - 2 , ln( x + 1) is undefined.
ln( x + 1) = ln(3x + 1) − ln x ⇒ ln( x + 1) = ln
⇒ x = 1+ 2
34.
(log x) 3 = log x 4 ⇒ (log x) 3 = 4 log x ⇒ let a = log x ⇒ a 3 − 4a = 0
⇒ a(a 2 − 4) = a(a − 2)(a + 2) = 0 ⇒ a = log x = 0, 2, or, − 2 ⇒ x = 10 0 = 1, 10 2 = 100, 10 − 2 = 0.01
38.
3log x = 3 x ⇒ log 3log x = log 3 x ⇒ log x log 3 = log 3 x = log 3 + log x
log 3
log 3
⇒ (log 3 − 1) log x = log 3 ⇒ log x =
⇒ x = 10 log 3−1
log 3 − 1
Evaluate to four decimal digits.
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 14
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44. log 4 23 =
log 23
≈ 2.262
log 4
77.Compound Interest: How many years, to the nearest year, will it take a sum of money
to double if it is invested at 15% compounded annually?

Use the formula A = P1 +

nt
r
 .
n
1t
t
115
log 2
 115 
 0.15 
2 P = P 1 +
= log 2 ⇒ t =
≈ 4.96 ≈ 5 years
 ⇒ t log
 ⇒2=
115
1 
100
 100 

log
100
83.World Population: A mathematical model for world population growth over short
periods of time is given by P = P0 e rt , where P is the population after t years,
P0 is the population at t = 0, and the population is assumed to grow continuously at the
annual rate r. How many years, to the nearest year, will it take the world population to be
double if it grows continuously at an annual rate of 2%?
ln 2
2 P0 = P0 e 0.02t ⇒ 2 = e 0.02t ⇒ ln 2 = 0.02t ⇒ t =
≈ 35 years
0.02
60.
Bacterial Growth. If bacteria in a certain culture double every ½ hour, write an equation
that gives the number of bacteria N in the culture after t hours, assuming the culture has
100 bacteria at he start. Graph the equation for 0 ≤ t ≤ 5 .
Since it doubles every ½ hour,
N = 100(2 2t ),0 ≤ t ≤ 5
And, the graph is as follows:
66.Finance. Suppose $2500 is invested at 7% compounded quarterly. How much money
will be in the account in:
( A ) ¾ year?
nt
 r
 0.07 
A = P1 +  = 25001 +

4 
 n

( B ) 15 years?
4(3 / 4)
= $2633.56
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 14
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
A = P1 +

nt
r
 0.07 
 = 25001 +

n
4 

4 (15 )
= $7079.54
44.Population Growth: If the population in Mexico is around 100 million people now and
if the population grows continuously at an annual rate of 2.3%, what will the population
be in 8 years? Compute the answer to 2 significant digits.
A = Pe rt = 100e ( 0.023)8 million ≈ 120,201,582 ≈ 120millions
54.
Money Growth.
Barron’s, a national business and financial weekly, published the following “ Top
Savings Deposit Yields” for 2 and ½ -year certificate of deposit accounts:
Gill Saving
8.30%(CC)
Richardson Savings and Loan
8.40%(CQ)
USA Savings
8.25%(CD)
In another issue of Barron’s, 1-year certificate of deposit accounts included:
Alamo Savings
Lamar Savings
8.25%(CQ)
8.05%(CC)
Compute the value of $10,000 invested in each account at the end of 1 year.
4
 0.0825 
Answer: CQ (compounded quarterly): A = 100001 +
 ≈ $10,850.88
4 

CC (compounded continuously): A = 10000e 0.0805 ≈ $10838.29
56.Present Value: A promissory note will pay $50,000 at maturity 5.5 years from now.
How much should you be willing to pay for the note now if the note gains value at a rate
of 10 % compounded continuously?
50000 = Pe ( 0.1)(5.5) ⇒ P =
50000
≈ $28,847.49
e ( 0.1)(5.5)
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 14
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College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 14
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