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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!)