Supplement for Exponential and Logarithmic Equations and Applications 
M122 College Algebra Section 3.3 

Name___________________________________ 

Solve the equation algebraically. 17) ln(y -6) + ln 4 = 2 

1) 5x = 
1 

625 
18) -2ln(x-5)-3= 11 

2) 2x = 16 Solve the problem. 

19) During 1991, 200,000 people visited Rave 

3) 4(3x -1) = 12 Amusement Park. During 1997, the number 
had grown to 834,000. If the number of visitors 
to the park obeys the law of uninhibited 

4) 2e4x -6 = 6 
growth, find the exponential growth function 
that models this data. 
5) 5+3 e5t-4 = 65 

20) A culture of bacteria obeys the law of 

6) 3 · 52t -1 = 75 uninhibited growth. If 140,000 bacteria are 
present initially and there are 609,000 after 6 
hours, how long will it take for the population 

7) (13)
x 
= 16 to reach one million? 

21) The half-life of silicon-32 is 710 years. If 70 
8) 71x = 60 grams is present now, how much will be 
present in 900 years? (Round your answer to 
9) lnx =-9 three decimal places.) 

10) log2 x = 3 22) A bacterial culture has an initial population of 
10,000. If its population declines to 3000 in 2 
hours, what will it be at the end of 4 hours? 


8

11) log


= 3

x 

27 
23) In 1985, the average annual consumption of 
beef B was about 79 lbs per person. In 1996, it 

12) log 4x = log 3 + log (x -2) was about 66 lbs per person. Assuming 
consumption is decreasing according to the 
exponential-decay model, write an equation 

13) log2(3x -2) -log2(x -5) = 4 that describes beef consumption t years after 
1985. 

14) log (4 + x) -log (x -3) = log 2 
24) The number of farms in a certain state has 
15) 9 -log3(x + 8) = 8 declined continually since 1950. In 1950, there 
were 92,120 farms and in 1995 that number 
had decreased to 27,339. Assuming the 
16) 9ln(x -3) = 1 number of farms decreased according to the 
exponential model, find the value of k and 
predict the number of farms in 2005. 

1 


Answer Key 
Testname: SOLVING EXP AND LOG EQUATIONS 

1) {-4} 

2) {4}
ln12
ln4 +1


3) or approx. 0.93 

3 

4) ln(3)+6 or approx. {1.775} 

4 

5) ln(20)+4 or approx. {1.40} 

5 

6) t = 
3 

2 

7) ln16 or approx. {-2.52} 

ln(1/3) 

8) 0.9605 

9) x = 
e19 

10) {8} 

11) {2}

3 

12) No solution. (x=-6 is not in the domain)
13) x = 6
14) {10}
15) x =-5


16) x = e1/9 + 3 

17) e2+24 

4 

18) x= 
e17 +5 

19) f(t) = 200,000e0.238t
20) 8.024 hours
21) 29.074
22) 900


23) P(t) = 79e-0.016t
24) k =-0.027 ; 20,865 farms