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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:
loga√a6b8a2b5.
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
x−2xlog2(5)=−1.
Factor out x on the left to get
x(1−2log2(5))=−1.
Now divide by 1−2log2(5) to get
x=−11−2log2(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)=2−3.
Now on the left use the inverse property and we get
x=2−3.
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(x−4)=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(xx−4)=ln(3).
Now plug both sides into the function ex to get
eln(xx−4)=eln(3).
Use the inverse property on each side to get
xx−4=3.
Now mulitply by x−4 (keep in mind we are forced to take x−4≠0) to get
x=3(x−4).
Distribute the 3 on the right to get
x=3x−12.
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+3.75e−0.32t.
Find the population after 0,1,5,10,15, and 20 months.
Solution: For this problem, you just compute
P(0)=30401+3.75e−0.32(0),
P(1)=30401+3.75e−0.32(1),
P(5)=30401+3.75e−0.32(5),
P(10)=30401+3.75e−0.32(10),
P(15)=30401+3.75e−0.32(15),
and
P(20)=30401+3.75e−0.32(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!)