Logarithms Questions and Answers

Answer: B

The Laplace method runs in three steps:

1. Transform the ODE into an algebraic subsidiary equation in \( s \)
2. Solve that equation for \( Y(s) \) by algebra
3. Take the inverse transform to recover \( y(t) \)

The initial conditions enter at step 1, which is why no arbitrary constants remain at the end.

Answer: C

Express 9 as a power of 27 so the given logarithm can be reused. Since \( 27 = 3^{3} \) and \( 9 = 3^{2} \),

\[ \log 9 \;=\; \log 27^{\,2/3} \;=\; \tfrac{2}{3}\log 27 \;=\; \tfrac{2}{3}\left(1.431\right) \;=\; 0.954 \]

The step that does the work is the power rule, \( \log a^{n} = n\log a \). Nothing here needs a table: any logarithm of a power of 3 follows from \( \log 27 \) once the exponent is written as a fraction of 3.

Answer: B

Write the hyperbolic sine in exponential form and use the shift theorem.

\[ \sinh t \cos t = \frac{1}{2}\left(e^{t} - e^{-t}\right)\cos t \]

With \( \mathcal{L}\{e^{at}\cos t\} = \dfrac{s-a}{(s-a)^{2}+1} \):

\[ \mathcal{L} = \frac{1}{2}\left[\frac{s-1}{s^{2}-2s+2} - \frac{s+1}{s^{2}+2s+2}\right] \]

Over the common denominator \( (s^{2}-2s+2)(s^{2}+2s+2) = s^{4}+4 \):

\[ \text{numerator} = (s-1)(s^{2}+2s+2) - (s+1)(s^{2}-2s+2) = 2s^{2} - 4 \]

\[ \mathcal{L}\{\sinh t \cos t\} = \frac{s^{2}-2}{(s^{2}+2s+2)(s^{2}-2s+2)} \]

As a check, \( \sinh t \cos t \approx t \) for small \( t \), and the expression behaves like \( \tfrac{1}{s^{2}} \) for large \( s \) - which is \( \mathcal{L}\{t\} \).

Answer: C

Growth proportional to the amount present gives exponential growth, and doubling every week means the population multiplies by 2 each week.

\[ A(4) = A_{0} \times 2^{4} = 16\,A_{0} \]

The doublings compound rather than add: four weeks give \( 2 \times 2 \times 2 \times 2 = 16 \) times, not \( 2 \times 4 = 8 \) times.

Answer: C

The transfer function is the ratio of the transformed output to the transformed input, with zero initial conditions:

\[ H(s) = \frac{Y(s)}{X(s)} = \frac{\text{output}}{\text{input}} \]

It characterises the system independently of any particular input, which is what makes it so useful in control and signal analysis.

Answer: B

The word differential points to differentials and hence to derivatives, and equation to a statement of equality. A differential equation is therefore an equation containing derivatives of an unknown function.

\[ \frac{dy}{dx} + 2y = e^{x} \]

An equation containing integrals instead would be an integral equation.

Answer: D

Expand the argument first:

\[ \sin(\omega t - \theta) = \sin \omega t \cos\theta - \cos \omega t \sin\theta \]

The transform is linear, and \( \cos\theta \) and \( \sin\theta \) are constants, so each term transforms separately:

\[ \mathcal{L}\{\sin(\omega t - \theta)\} = \cos\theta\left(\frac{\omega}{s^{2}+\omega^{2}}\right) - \sin\theta\left(\frac{s}{s^{2}+\omega^{2}}\right) \]

The minus sign carries over from the expansion; a phase of \( +\theta \) would make it a plus.

Answer: A

"Proportional to the number present" translates directly into a constant of proportionality times \( A \):

\[ A' = \frac{dA}{dt} = cA \]

This is the exponential growth law, whose solution is \( A(t) = A_{0}e^{ct} \). The same equation governs radioactive decay, with \( c \) negative.

Answer: C

A logarithm to base \( e \), where \( e \approx 2.71828 \), is a natural logarithm, written \( \ln x \) or \( \log_e x \).

• Natural - base \( e \), the base at which the derivative of \( \log_b x \) is exactly \( 1/x \), which is why calculus uses it throughout.
• Common - base 10, the one meant by a bare \( \log \) in most arithmetic work.

There is no such thing as a pure or an infinite logarithm; both are invented distractors.

Answer: B

Start from Boyle's law at constant temperature:

\[ V = \frac{c}{p} \]

Differentiating with respect to \( p \):

\[ \frac{dV}{dp} = -\frac{c}{p^{2}} \]

Since \( c = pV \), this becomes

\[ \frac{dV}{dp} = -\frac{pV}{p^{2}} = -\frac{V}{p} \]

The negative sign reflects that volume falls as pressure rises.