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Problem A: Show that for $x>0$ and $n$ a positive integer that $\ln(x^n)=n\ln(x)$ using calculus.
Solution: Differentiate the left-hand side to get
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x^n) = \dfrac{1}{x^n} \dfrac{\mathrm{d}}{\mathrm{d}x} x^n = \dfrac{nx^{n-1}}{x^n} = \dfrac{n}{x}.$$
Differentiate the right-hand side to get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} n \ln(x) = \dfrac{n}{x}.
\end{array}$$
This shows that both $\ln(x^n)$ and $n\ln(x)$ are antiderivatives of the function $\dfrac{n}{x}$. Consequently, Theorem 4.1 (pg. 244) tells us that there is a constant $C$ so that for all valid $x$,
$$\ln(x^n) = n\ln(x) + C.$$
What is this constant $C$? One way to find out is to let $x=1$ in this formula to see
$$\ln(1^n) = n\ln(1)+C,$$
which simplifies to
$$0=0+C,$$
hence $C=0$. Therefore we have shown that
$$\ln(x^n) = n\ln(x),$$
as desired.
Section 5.1 #108: Use the approximation (known as the prime number theorem) $p(x) \approx \dfrac{x}{\ln(x)}$, where $p(x)$ denotes the prime counting function, to esimate the number of prime numbers less than or equal to...
(a) $x=1000$,
(b) $x=1000000$, and
(c) $x=1000000000$.
Solution: Simply plug these values into the approximation. For $x=1000$ we get
$$p(1000) \approx \dfrac{1000}{\ln(1000)} = 144.764\ldots,$$
for $x=10000$ we get
$$p(10000) \approx \dfrac{10000}{\ln(10000)} = 1085.736\ldots,$$
and for $x=1000000000$ we get
$$p(1000000000) \approx \dfrac{1000000000}{\ln(1000000000)} = 48254942.433\ldots$$
Section 5.2 #97: A population of bacteria $P$ is changing at a rate of
$$\dfrac{\mathrm{d}P}{\mathrm{d}t} = \dfrac{3000}{1+0.25t},$$
where $t$ is measured in days. The initial population (i.e. $t=0$) is $1000$. Write an equation that gives the population at any time $t$, then find the population when $t=3$ days.
Solution: To say the initial population is $1000$ means $P(0)=1000$. Recall the fundamental theorem of calculus shows that $\displaystyle\int_a^b \dfrac{\mathrm{d}f}{\mathrm{d}x} \mathrm{d}x = f(b)-f(a)$. We will proceed by integrating each side of the given equality from $0$ to $x$ (with $x>0$). The left-hand side of the equality becomes
$$\displaystyle\int_0^x \dfrac{\mathrm{d}P}{\mathrm{d}t} \mathrm{d}t = P(x)-P(0)=P(x)-1000$$
while the right-hand side becomes, using $u=1+0.25t$ hence $\mathrm{d}u=0.25 \mathrm{d}t$ or equivalently $4\mathrm{d}u=\mathrm{d}t$, yielding
$$\displaystyle\int_0^x \dfrac{3000}{1+0.25t} \mathrm{d}t = (4)(3000) \displaystyle\int_1^{1+0.25x} \dfrac{1}{u} \mathrm{d}u = 12000 \ln(|1+0.25x|).$$
Therefore combining the two sides we see
$$P(x) = 1000 + 12000 \ln(|1+0.25x|).$$
At time $x=3$ days, the population is $P(3)$ which yields a population of
$$P(3) = 7715.39\ldots \approx 7715$$
bacteria.
Section 5.2 #108: Prove that the function
$$F(x)=\displaystyle\int_x^{2x} \dfrac{1}{t} \mathrm{d}t$$
is constant for $x$ in the interval $(0, \infty)$.
Solution: We will interpret this function in terms of logarithms. Recall the property of logarithms that for any valid $c$, $\displaystyle\int_a^b f(t) \mathrm{d}t = \displaystyle\int_a^c f(t) \mathrm{d}t + \displaystyle\int_c^b f(t) \mathrm{d}t$ and the property that $\displaystyle\int_a^b f(t) \mathrm{d}t=-\displaystyle\int_b^a f(t) \mathrm{d}t$. Using $c=1$ and the definition of $\ln$, we see
$$\begin{array}{ll}
F(x) &= \displaystyle\int_x^{2x} \dfrac{1}{t} \mathrm{d}t \\
&= \displaystyle\int_x^1 \dfrac{1}{t} \mathrm{d}t + \displaystyle\int_1^{2x} \dfrac{1}{t} \mathrm{d}t \\
&= -\displaystyle\int_1^x \dfrac{1}{t} \mathrm{d}t + \displaystyle\int_1^{2x} \dfrac{1}{t} \mathrm{d}t \\
&=-\ln(x) + \ln(2x).
\end{array}$$
Using the properties of logarithms (Theorem 5.2, pg. 319) this shows that
$$F(x) = \ln(2x)-\ln(x) = \ln \left( \dfrac{2x}{x} \right) = \ln(2),$$
which is a constant, as was to be shown.
