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Using Logarithms to solve Bx = A
1. Given the exponential equation Bx = A, use the change of base formula to solve for x. x = lnA/lnB
2. Given the logarithmic equation; log B A  x , use the change of base formula to solve for x. x = lnA/lnB
Solve:
3. 25x = 125
x
log 125 3

log 25 2
6. 4x = 100
x
log 100
 3.32
log 4
5. 36x = 216
x  4 12.8  1.891
x
7. 5x = .04
8. 10x = 125
x
9. 2x-2 = 128
x2
4. x4 = 12.8
ln 0.04
 2
ln 5
x  log 125; x  2.097
10. 8x+3 = 500
log 128
;x  9
log 2
12. 34x+1 = 85
log 85
4x  1 
; x  .761
log 3
x
x3
11. 252x = 125
log 500
; x  0.011
log 8
2 x  40
6 x 1  216
x
log 40
; x  5.322
log 2
x
1.08 x  12.5
1.025 x  8 / 13
 .18 
18. 1501 

12 

1.01512 x  37 / 3
12 x 
12 x
 1850
log 37 / 3
; x  14.062
log 1.015
log 125
; x  0.75
log 25
14. 4(6)x-1 = 864
16. 650(1.025) = 400
log 12.5
; x  32.32
log 1.08
2x 
13. 3(2)x = 120
15.b 100(1.08) = 1250
x
log 216 3

log 36 2
x
log 8 / 13
; x  19.662
log 1.025
 .25 
19. 90001 

4 

.9375 4 x  5 / 6
4x 
4x
x 1 
log 216
;x  4
log 6
 .04 
17. 3001 

2 

1.02 2 x  2.5
2x 
2x
 750
log 2.5
; x  23.136
log 1.02
52 x
 1.08 
 7500 20. 35001 
  20000
52 

1327 / 1300 52 x  40 / 7
log 5 / 6
; x  .706
log .9375
52 x 
log 40 / 7
; x  1.631
log 1327 / 1300
Properties of Logarithms
A. Rewrite as a single logarithm
B. Evaluate your answer from part A
21) log x  4 log 2
1
22) log 5 u  log 5 8
3
log 16  1.204
log 5 2  0.431
23) log m  log 3  log 7
24) log y  log 35  log 7
log 21  1.322
log 5  0.699
25) log p  log 4  3 log 5
26) 4 log x  log 32  log 2
log x 4  log 16
log 4  (log 5 )
3
x 4  16, x  2
log 500  2.699
log 16  2
27) log 2 x  2 log 6  log 3
28) log 2
t
 2 log 2 5  3 log 2 2
3
t
 log 2 25  log 2 8
3
t
log 2  log 2 200
3
t
 200;
3
t  600
log 2
log 2 x  log 36  log 3
2 x  108;
x  54
Write each number as a decimal (do as many as possible without a calculator)
29) ln e14  14
31) 5 ln e  5
32) 5 ln e 2  10
33) log 18 18 20  y
34) 7 log 3 3  8 log 3 3  y
34) log 8 100
log 3 37  log 3 38  log 3
y = 20
37
38
1
8
log 100  log 100
1
 1 or,
1
1
1
3
2 8
 log 10
 log 10 4 
7 log 3 3  8log 3 3  7(1)  8(1)
4
log 3
8
 
Write in exponential form
33) ln 42  3.738
34) ln 2.4  0.875
35) ln e3  y
e 3.738  42
e 0.875  2.4
e y  e3
Write in logarithmic form
36) e 7  1097
37) e x  5.755
38) 2e x  12.8
ln 1097  7
ln 5.755  x
ln 6.4  x or ln12.8  x 2
39) Danny’s college savings are invested in a bond that pays an annual interest rate of 6.2% compounded
continuously. How long will it take the money to triple?
3P  Pe0.062t   3  e0.062t   ln 3  .062t   t 
ln 3
  t  17.720 yrs
.062
40) At what rate would you have to invest your money for it to double if it were compounded continuously for 8
years?
2  e 8 r   ln 2  8r   r 
ln 2
  rate  8.7%
8
41) At what rate would you have to invest your money for it to double if it were compounded monthly for 2 years?

24

r 
r
r

2  1      24 2  1     24 2  1    12 24 2  1  35.16%
12
12
 12 
42) How long will it take $200 to grow to $1000 if it was invested @ an annual rate 8% compounded quarterly?
4t
ln 5
1  ln 5 
 .08 
1000  2001     5  1.02 4t  
 4t   t  
  t  20.318 yrs
4 
ln 1.02
4  ln 1.02 

43) When a certain medical drug is administered to a patient, the number of milligrams remaining in the patient’s
bloodstream after t hours is modeled by D(t) = 50e-0.2t
A) Create a word model to describe the function D(t)
50 mg are injected and the drug decreases exponentially by 20% every hour
B) How many milligrams would remain in the blood after 3 hours? D(t) = 50e-0.2(3) = 27.44 mg
C) Due to the toxicity of the drug and the possibility negative side effects, patients should not take another dose
until only 10% of the drug remains in the system. Approximately how long after the first dose should you wait until
taking a second dose?
5 = 50e-0.2(t)  0.1 = e-0.2(t) ln0.1 = -0.2t  t = ln0.1/-.02  -5ln0.1 about 11.51 years
44) Nancy wants to invest $4000 in savings certificates that bear an interest rate of 5.75% compounded semiannually. How long should she invest for if she wants to make $1200 in interest?
2t
2t
ln 1.3
 0.0575 
 0.0575 
5200  40001 
t  4.628 years
    1.3  1 
    2t 
2 
2 
ln 1.08875


45) Radioactive iodine decays exponentially in mass by 8.7% every day. If a sample began with 20 grams, how
long would it take until there were 10 grams remaining?
10 = 20e-0.087t 0.5 = e-0.087t  t = ln0.5/-0.87: about 7.96 years
46) On January 1st 2002 the population of a country had a population of 4,100,000 people. Since that time it has
been growing at an annual rate of 3.5% (this is not continuous/exponential). Assume that this has been and always
will remain the same.
A) What was the population of the country 6 months later? 4100000(1.035)(0.5): about 4,171,133
B) What was the population of the country 9 years later? 4100000(1.035)(9): about 5,587,879
C) What will be the population of the country two decades later? 4100000(1.035)(20): about 5,8158,134
D) According to this model what would the population have been in 1970? 4100000(1.035)(-32): about 1,363,618
47) Find the missing variable and create a word problem to match each problem.
A) 4200 = 9000e15r
$9000 decreases exponentially. What is the rate that leaves you with $4200 after 15 years?
B) 4200 = 9000(r)15
$9000 decreases annually. What is the rate that leaves you with $4200 after 15 years?
r 

C) 42001  
 12 
180
$4200 is compounded monthly for 15 years at an unknown rate.