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**Section 5.4 #23**: Express as a sum or difference of logarithms:
$$\log_a(6xy^5z^4).$$
*Solution:* We use the rule that the log of the product is the sum of logs to write
$$\log_a(6xy^5z^4)=\log_a(6)+\log_a(x)+\log_a(y^5)+\log_a(z^4).$$
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
$$\log_a(6xy^5z^4)=\log_a(6) + \log_a(x) + 5\log_a(y) + 4\log_a(z).$$

**Section 5.4 #34:** Express as a sum or difference of logarithms:

$$\log_a \sqrt{ \dfrac{a^6b^8}{a^2b^5} }.$$
*Solution:* First simplify the inside of the square root:
$$\log_a \left( \sqrt{ \dfrac{a^6b^8}{a^2b^5} } \right) = \log_a \left( \sqrt{ a^4 b^3} \right).$$
Now use the rule that the log of the power is the multiple of the log to get
$$\log_a \left( \sqrt{a^4 b^3} \right) = \log_a \left( \left( a^4 b^3 \right)^{\frac{1}{2}} \right) = \dfrac{1}{2} \log_a (a^4 b^3).$$
Now use the rule that the log of the product is the sum of the logs:
$$\dfrac{1}{2} \log_a(a^4 b^3) = \dfrac{1}{2} \left[ \log_a(a^4) + \log_a(b^3) \right].$$
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:
$$\dfrac{1}{2} [ \log_a(a^4) + \log_a(b^3) ] = \dfrac{1}{2} [ 4 + 3 \log_a(b) ].$$
**Section 5.4 #41:** Express as a single logarithm:

$$\dfrac{1}{2} \log_a(x) + 4\log_a(y) - 3\log_a(x).$$
*Solution:* First use the multiple of a log is the log of a power to write
$$\dfrac{1}{2} \log_a(x) + 4\log_a(y) - 3\log_a(x) = \log_a(x^{\frac{1}{2}}) + \log_a(y^4) - \log_a(x^3).$$
Now use the sum of logs is the log of the product to write
$$\log_a(x^{\frac{1}{2}})+ \log_a(y^4) - \log_a(x^3) = \log_a(x^{\frac{1}{2}} y^4) - \log_a(x^3).$$
Now use the difference of logs is the quotient of logs to write
$$\log_a \left( x^{\frac{1}{2}} y^4 \right) - \log_a(x^3) = \log_a \left( \dfrac{x^{\frac{1}{2}}y^4}{x^3} \right) = \log_a \left( \dfrac{y^4}{x^{\frac{5}{2}}} \right).$$

**Section 5.4 #69:** Simplify
$$3^{\log_3(4x)}.$$

*Solution:* Using the inverse property, we get
$$3^{\log_3(4x)}=4x.$$

**Section 5.5 #1:** Solve
$$3^x = 81.$$

*Solution:* We want to isolate $x$. To do that we will take $\log_3$ of both sides to remove the base of the exponential:
$$\log_3(3^x) = \log_3(81).$$
This gives us
$$x = \log_3(81) = \log_3(3^4) = 4.$$

**Section 5.5 #20:** Solve
$$1000e^{0.09t} = 5000.$$

*Solution:* We want to solve for $t$. To do that, first divide both sides by $1000$ to get
$$e^{0.09t} = \dfrac{5000}{1000} = 5.$$
Now to remove the base of the exponential, take $\ln$ of both sides to get
$$\ln \left( e^{0.09t} \right) = \ln(5).$$
Now on the left we get simplification and see
$$0.09t = \ln(5).$$
To find $t$ now just divide by $0.09$ to get
$$y = \dfrac{\ln(5)}{0.09}.$$

**Section 5.5 #28:** Solve
$$2^{x+1}=5^{2x}.$$
*Solution:* There is no "better" base of either exponential to remove -- just pick one. We will remove the $2$. Take $\log_2$ of both sides to get
$$\log_2 \left( 2^{x+1} \right) = \log_2 \left( 5^{2x} \right).$$
Using the inverse property on the left and the log of the power is the multiple of log property on the right we get
$$x+1 = 2x\log_2(5).$$
At this point we want to solve for $x$ (it's just a linear equation!) and so subtract $2x\log_2(5)$ from both sides and subtract $1$ from both sides to get
$$x-2x\log_2(5) = -1.$$
Factor out $x$ on the left to get
$$x(1-2\log_2(5))=-1.$$
Now divide by $1-2\log_2(5)$ to get
$$x = \dfrac{-1}{1-2\log_2(5)}.$$

**Section 5.5 #32:** Solve
$$\log_2(x) = -3.$$
*Solution:* We would like to recover the $x$ on the left-hand side. To do this plug both sides into the function $2^x$ to get
$$2^{\log_2(x)} = 2^{-3}.$$
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 $e^x$ to get
$$e^{\ln(x)}=e^1.$$
Therefore
$$x = e^1 = e.$$

**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
$$\ln \left( \dfrac{x}{x-4} \right) = \ln(3).$$
Now plug both sides into the function $e^x$ to get
$$e^{\ln \left( \frac{x}{x-4} \right)} = e^{\ln(3)}.$$
Use the inverse property on each side to get
$$\dfrac{x}{x-4} = 3.$$
Now mulitply by $x-4$ (keep in mind we are forced to take $x-4 \neq 0$) 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
$$x = \dfrac{12}{2} = 6.$$

(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
$$P(t) = 100e^{0.117t}.$$
For (b) we compute
$$P(7) = 100e^{0.117(7)}$$
and for the "2 weeks", keep in mind the variable $t$ is measured in days, so we write
$$P(14)=100e^{0.117(14)}.$$
For (c) we must find the doubling time, i.e. the value of $t$ that satisfies
$$2(100) = 100e^{0.117t}.$$
Therefore
$$2 = e^{0.117t},$$
and so
$$\ln(2) = 0.117t.$$
Therefore
$$t = \dfrac{\ln(2)}{0.117}.$$

**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
$$P(t) = \dfrac{3040}{1+3.75e^{-0.32t}}.$$
Find the population after $0,1,5,10,15,$ and $20$ months.

*Solution:* For this problem, you just compute
$$P(0) = \dfrac{3040}{1+3.75e^{-0.32(0)}},$$
$$P(1) = \dfrac{3040}{1+3.75e^{-0.32(1)}},$$
$$P(5) = \dfrac{3040}{1+3.75e^{-0.32(5)}},$$
$$P(10) = \dfrac{3040}{1+3.75e^{-0.32(10)}},$$
$$P(15) = \dfrac{3040}{1+3.75e^{-0.32(15)}},$$
and
$$P(20) = \dfrac{3040}{1+3.75e^{-0.32(20)}}.$$
(note: you may use a calculator to find the decimal values for these expressions, but you won't have that ability on the exam!)