Processing math: 100%
AMPS | THARC | KE8QZC | SFW | TSW | WW
ORCID iD icon

Back to the class
Section 5.4 #23: Express as a sum or difference of logarithms: loga(6xy5z4). Solution: We use the rule that the log of the product is the sum of logs to write loga(6xy5z4)=loga(6)+loga(x)+loga(y5)+loga(z4). Now we apply the rule that the log of the power is the multiple of the log in the third and fourth terms to get loga(6xy5z4)=loga(6)+loga(x)+5loga(y)+4loga(z).

Section 5.4 #34: Express as a sum or difference of logarithms:
logaa6b8a2b5. Solution: First simplify the inside of the square root: loga(a6b8a2b5)=loga(a4b3). Now use the rule that the log of the power is the multiple of the log to get loga(a4b3)=loga((a4b3)12)=12loga(a4b3). Now use the rule that the log of the product is the sum of the logs: 12loga(a4b3)=12[loga(a4)+loga(b3)]. We can simplify the first logarithm by using the inverse property and the second one by using the log of the power is the multiple of the log property: 12[loga(a4)+loga(b3)]=12[4+3loga(b)]. Section 5.4 #41: Express as a single logarithm:
12loga(x)+4loga(y)3loga(x). Solution: First use the multiple of a log is the log of a power to write 12loga(x)+4loga(y)3loga(x)=loga(x12)+loga(y4)loga(x3). Now use the sum of logs is the log of the product to write loga(x12)+loga(y4)loga(x3)=loga(x12y4)loga(x3). Now use the difference of logs is the quotient of logs to write loga(x12y4)loga(x3)=loga(x12y4x3)=loga(y4x52).

Section 5.4 #69: Simplify 3log3(4x).
Solution: Using the inverse property, we get 3log3(4x)=4x.

Section 5.5 #1: Solve 3x=81.
Solution: We want to isolate x. To do that we will take log3 of both sides to remove the base of the exponential: log3(3x)=log3(81). This gives us x=log3(81)=log3(34)=4.

Section 5.5 #20: Solve 1000e0.09t=5000.
Solution: We want to solve for t. To do that, first divide both sides by 1000 to get e0.09t=50001000=5. Now to remove the base of the exponential, take ln of both sides to get ln(e0.09t)=ln(5). Now on the left we get simplification and see 0.09t=ln(5). To find t now just divide by 0.09 to get y=ln(5)0.09.

Section 5.5 #28: Solve 2x+1=52x. Solution: There is no "better" base of either exponential to remove -- just pick one. We will remove the 2. Take log2 of both sides to get log2(2x+1)=log2(52x). Using the inverse property on the left and the log of the power is the multiple of log property on the right we get x+1=2xlog2(5). At this point we want to solve for x (it's just a linear equation!) and so subtract 2xlog2(5) from both sides and subtract 1 from both sides to get x2xlog2(5)=1. Factor out x on the left to get x(12log2(5))=1. Now divide by 12log2(5) to get x=112log2(5).

Section 5.5 #32: Solve log2(x)=3. Solution: We would like to recover the x on the left-hand side. To do this plug both sides into the function 2x to get 2log2(x)=23. Now on the left use the inverse property and we get x=23.

Section 5.5 #35: Solve ln(x)=1. Solution: We would like to recover the x on the left-hand side. To do this plug both sides into the function ex to get eln(x)=e1. Therefore x=e1=e.

Section 5.5 #50: Solve ln(x)ln(x4)=ln(3). Solution: We need to get a single ln on each side in order to cancel it. So on the left use the difference of logs is the log of the quotient to get ln(xx4)=ln(3). Now plug both sides into the function ex to get eln(xx4)=eln(3). Use the inverse property on each side to get xx4=3. Now mulitply by x4 (keep in mind we are forced to take x40) to get x=3(x4). Distribute the 3 on the right to get x=3x12. Therefore add 12 and subtract by x to get 12=2x, and finally divide by 2 to see x=122=6.

(note: you can use a calculator to make these values into decimals, but that won't be possible on the exam!!)

Section 5.6 #2: Under ideal conditions, a population of rabbits has an exponential growth rate of 11.7% per day. Consider an initial population of 100 rabbits.
a.) Find the exponential growth function.
b.) What will the population be after 7 days? after 2 weeks?
c.) Find the doubling time.
Solution: For (a), first note that 11.7%=0.117. So the requested exponential function is P(t)=100e0.117t. For (b) we compute P(7)=100e0.117(7) and for the "2 weeks", keep in mind the variable t is measured in days, so we write P(14)=100e0.117(14). For (c) we must find the doubling time, i.e. the value of t that satisfies 2(100)=100e0.117t. Therefore 2=e0.117t, and so ln(2)=0.117t. Therefore t=ln(2)0.117.

Section 5.6 #18: A lake is stocked with 640 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict the growth of that type of fish in the lake to a limiting value of 3040. The population of fish in the lake after time t, in months, is given by the function P(t)=30401+3.75e0.32t. Find the population after 0,1,5,10,15, and 20 months.
Solution: For this problem, you just compute P(0)=30401+3.75e0.32(0), P(1)=30401+3.75e0.32(1), P(5)=30401+3.75e0.32(5), P(10)=30401+3.75e0.32(10), P(15)=30401+3.75e0.32(15), and P(20)=30401+3.75e0.32(20). (note: you may use a calculator to find the decimal values for these expressions, but you won't have that ability on the exam!)