Logarithms Questions and Answers

Answer: C

The value of the expression is 1.

Change every logarithm to a common base: \(\log_{u} vw = \dfrac{\log v + \log w}{\log u}\). Then

\[ \begin{aligned} S &= \log u + \log v + \log w \\ 1 + \log_{u} vw &= \frac{S}{\log u} \\ \frac{1}{1 + \log_{u} vw} &= \frac{\log u}{S} \end{aligned} \]

The other two terms give \(\log v\) and \(\log w\) over the same denominator S, so the three add up to

\[ \frac{\log u + \log v + \log w}{S} = \frac{S}{S} = 1 \]

The result does not depend on u, v or w, so -1, 0 and 3 cannot be right.

Answer: C

Prime notation is convenient only for low orders:

\[ y', \quad y'', \quad y''' \]

Beyond the third derivative the primes become hard to count, and the numeral form \( y^{(4)} \) or the Leibniz form \( \tfrac{d^{4}y}{dx^{4}} \) is used instead.

Answer: B

The notation \( \ln \) always means the logarithm to base \( e \), so \( \ln 2 \) is a natural logarithm - the power to which \( e \) must be raised to give 2:

\[ \ln 2 \;=\; \log_e 2 \;\approx\; 0.693 \]

The trap is the numeral 2 in the bracket, which suggests base 2. The 2 is the argument, not the base. A binary logarithm is written \( \log_2 x \) and takes 2 as its base; the two are different, and indeed \( \log_2 2 = 1 \) whereas \( \ln 2 \approx 0.693 \).

Answer: D

The standard transform pair is

\[ \mathcal{L}\{\sin \omega t\} = \frac{\omega}{s^{2}+\omega^{2}} \]

while its companion is

\[ \mathcal{L}\{\cos \omega t\} = \frac{s}{s^{2}+\omega^{2}} \]

The two differ only in the numerator: \( \omega \) for the sine, \( s \) for the cosine. That is the pair most often confused.

Answer: A

A deterministic model is one in which the present state is fixed entirely by the parameters and the earlier states: run it twice from the same start and it gives the same answer twice.

A stochastic or probabilistic model includes randomness, so repeated runs can differ.

Answer: D

Volume changes at the difference between what flows in and what flows out.

\[ \frac{dV}{dt} = \text{rate in} - \text{rate out} \]

The two rates are equal and constant, so the difference is zero:

\[ \frac{dV}{dt} = 0 \]

The volume is therefore constant - the tank is in steady state, neither filling nor draining.

Answer: D

The value prescribed at the starting point of a problem is the initial condition, also called the seed value.

\[ y(t_{0}) = y_{0} \]

It is what picks a single solution out of the general family, each member of which differs only by the arbitrary constant.

Answer: A

Boyle's law states that at constant temperature the pressure and volume of a gas are inversely proportional:

\[ pV = \text{constant} \quad\Longrightarrow\quad V = \frac{\text{constant}}{p} \]

Halving the pressure therefore doubles the volume. The approximation holds best at low pressure, where the gas is close to ideal.

Answer: D

A linear accelerator drives particles forward with oscillating electric fields, and an electric field can only exert force on something carrying charge.

\[ F = qE \]

So only charged particles - electrons, protons, ions - can be accelerated. A neutral particle has \( q = 0 \) and feels no force at all.

Answer: A

Napier's logarithms are the natural logarithms, taken to base \( e \) and written \( \ln x \).

Common logarithms, to base 10, came slightly later and are named after Briggs. The mantissa is only the fractional part of a logarithm, and the antilogarithm is the inverse operation, so neither names a system of logarithms at all.