MATH 1314 – TEST #3 REVIEW – CHAPTER 4
1. Graph f(x) and g(x) on the same grid. Use transformations of f(x) to get g(x).
Determine the domain and range for g(x).
a)
b)
c)
d)
e)
f)
f(x) = 2x g(x) = 2x–1
f(x) = 2x g(x) = 2x – 1
f(x) = 2x g(x) = –2x
f(x) = log2 x
g(x) = log2 (x – 2)
f(x) = log2 x
g(x) = log2 x – 1
f(x) = log2 x
g(x) = log2 (–x)
2. Suppose that you have $5000 to invest. Which investment yields the greater return over 5 years?
Investment A: 5.5% compounded semiannually
Investment B: 5.25% compounded monthly
3. Write each equation in exponential form:
a) log49 7 = ½
b) log2 x = 3
c) log3 81 = y
4. Write each equation in logarithmic form:
a) 63 = 216
b) b4 = 625
c) 13y = 874
5. Evaluate each expression without using a calculator (using log properties):
a) log4 64
b) log17 17
c) log3 38
d) ln e5
6. Find the domain of each logarithmic function. Write your answers in interval notation.
a) f(x) = log8 (x + 5)
b) f(x) = log (3 – x)
7. Use properties of logs to expand each log expression as much as possible:
a) log6 (36x3)
 x
b) log4 

 64 
 x y2 
c) log2 

 z 
d) ln
3
x
e
8. Use properties of logs to condense each log expression as much as possible:
a) log3 7 + log3 3
b) log 5 – 5 log x
c) 3 ln x + 4 ln y
d) ½ ln x – ln y
9. Use the change of base formula to evaluate the following to 4 decimal places:
a) log6 72348
b) log4 0.863
10. Solve each exponential equation. Give exact answers (no decimals!)
a) 24x–2 = 64
b) 3x+2 = 27–x
c) 8x = 12143
d) 9e5x = 1269
11. Solve each logarithmic equation. Give exact answers (no decimals!)
a) log4 (3x – 5) = 3
b) 4 ln(2x) + 3 = 15
c) log2 (x + 3) + log2 (x – 3) = 4
12. How long will it take $50,000 to triple in value at 7.5% annual interest compounded continuously?
(round your answer to the nearest tenth of a year)
13. What interest rate is required to triple an investment in 5 years if the interest is compounded
continuously? (round your answer to the nearest whole percent)
14. According to the US Bureau of the Census in 1990 there were 22.4 million residents of Hispanic
origin living in the US. By 2000 the number had increased to 35.3 million. The exponential growth
function A = 22.4 ekt describes the US Hispanic population, A, in million, t years after 1990.
a) Find k
b) Write the complete growth equation (ie, fill in k)
c) What will the Hispanic population be in 2010?
d) In which year will the Hispanic population reach 61 million?
15. The half-life of polonium-210 is 140 days. How long will it take for a sample of this substance to
decay to 20% of its original amount? (round your answer to the nearest whole day)
Answers:
a) f(x) = 2x
1.
g(x) = 2x–1 (right 1)
g(x) = 2x – 1 (down 1)
b) f(x) = 2x



























Domain of g(x): (–∞, ∞)
Range of g(x): (0, ∞)


Domain of g(x): (–∞, ∞)
Range of g(x): (–1, ∞)







g(x) = 2–x (reflect y-axis)
c) f(x) = 2x










g(x) = log2 (x – 2)
(right 2)
d) f(x) = log2 x









Domain of g(x): (–∞, ∞)
Range of g(x): (0, ∞)









Domain of g(x): (2, ∞)
Range of g(x): (–∞, ∞)








g(x) = log2 x – 1
(down 1)
e) f(x) = log2 x
f) f(x) = log2 x























Domain of g(x): (0, ∞)
Range of g(x): (–∞, ∞)













g(x) = log2 (–x)
(reflect y-axis)
Domain of g(x): (–∞, 0)
Range of g(x): (–∞, ∞)
2.
 $6558.26 (better investment)
Investment A: 5000 1  .055
2 
10
 $6497.16
Investment B: 5000 1  .0525
12 
60
3.
a) 49 ½ = 7
b) 23 = x
c) 3y = 81
6.
a) x > 5  (–5, ∞)
b) 3 – x > 0  x < 3  (–∞, 3)
7.
a)
b)
c)
d)
8.
a) log3 21
 5
b) log  5 
x 
c) ln (x3 y4)
 x
d) ln 

 y 
9.
a)
10.
a) 24 x 2  26  4 x  2  6  x  2
4.
a) log6 216 = 3
b) logb 625 = 4
c) log13 874 = y
5.
log6 36 + log6 x3 = 2 + 3 log6 x
log4 x – log464 = ½ log4 x – 3
log2 x + 2 log2 y – log2 z
1
/3 ln (x/e) = 1/3 (ln x – ln e) = 1/3 lnx – 1/3
log  72348 
 6.2448
log  6 
b) 3x 2   33 
x
b)
log .863 
 –0.1063
log  4 
 x  2  3x  x  21
c) ln  8 x   ln 12143   x ln8  ln12143  x 
ln12143
ln8
ln141
d) e 5 x  141  ln e 5 x  ln141  5x  ln141  x 
5
11.
a) 43  3x  5  x  23
e3
2
2
4
2
c) log2 (x  9)  4  2  x  9  25  x2  x  5  not  5
b) ln(2x)  3  e3  2x  x 
a)
b)
c)
d)
3
1
8
5
12.
150,000  50,000e.075t   3  e.075t   ln3  .075t  t  14.6 years
13.
3P  Pe5r   3  e5r   ln3  5r  r  22%
14.
a) 35.3  22.4e k 10  
353
 353 
 e10 k   ln 
  10k  k  .0454822005
224
 224 
b) f(t) = 22.4 e0.0454822005t
c) f(10) = 22.4 e(.0454822005×10) = 55.6 million
d) 61  22.4eKt 
15.
305  Kt 
 305 

e
 ln 
  Kt  t 
112
 112 
 305 
ln 

 112   22 years  Year 2012
K
1
1
1
Ao  Ao e k 140    e140 k   ln    140k  k  .0049510513
2
2
2
then f(t) = Ao e-.0049510513t
.2 Ao  Ao eKt   .2  eKt   ln .2   Kt  t 
ln(.2)
 325 days
K