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: A

The value is 79.

Combine the logarithms on the right: \(\log p + \log q + 2\log 3 = \log(9pq)\). Since log is one-to-one,

\[ p^{2} + q^{2} = 9pq \]

Square both sides and divide by \(p^{2}q^{2}\):

\[ \begin{aligned} p^{4} + 2p^{2}q^{2} + q^{4} &= 81p^{2}q^{2} \\ \frac{p^{4} + q^{4}}{p^{2}q^{2}} &= 81 - 2 = 79 \end{aligned} \]

81 forgets to remove the \(2p^{2}q^{2}\) cross term, 83 adds it instead of subtracting, and 9 is the unsquared ratio \((p^{2} + q^{2})/pq\).

Answer: A

The value is 79.

Combine the logarithms on the right: \(\log p + \log q + 2\log 3 = \log(9pq)\). Since log is one-to-one,

\[ \begin{aligned} p^{2} + q^{2} &= 9pq \\ (p^{2} + q^{2})^{2} &= 81p^{2}q^{2} \\ p^{4} + q^{4} &= 81p^{2}q^{2} - 2p^{2}q^{2} = 79p^{2}q^{2} \end{aligned} \]

So \((p^{4} + q^{4})/(p^{2}q^{2}) = 79\).

81 forgets to subtract the cross term \(2p^{2}q^{2}\) when squaring; 83 adds it instead; and 9 is the value of \((p^{2} + q^{2})/pq\), the first line, not the quantity asked for.

Answer: B

The load is 46.02.

Write the cycles to failure as \(N = Ae^{-kL}\). From the two data points:

\[ \begin{aligned} \frac{N(40)}{N(80)} &= e^{40k} = \frac{10000}{100} = 100 \\ k &= \frac{\ln 100}{40} \end{aligned} \]

For 5000 cycles, half of the 10000 at \(L = 40\):

\[ \begin{aligned} e^{-k(L - 40)} &= \tfrac{1}{2} \\ L &= 40 + \frac{40\ln 2}{\ln 100} = 40 + 6.02 = 46.02 \end{aligned} \]

5000 cycles lies between 10000 (at 40 units) and 100 (at 80 units), so the load must lie between 40 and 80: 40.00 itself gives 10000 cycles and 92.02, above 80 units, would give fewer than 100. 60.01 is far too high because the exponential falls fastest at first.

Answer: C

The value is 2.

The right-hand side is \(\tfrac{1}{2}\ln(xy) = \ln\sqrt{xy}\). Since ln is one-to-one, the condition says

\[ \frac{x + y}{2} = \sqrt{xy} \]

The arithmetic mean of two positive numbers equals their geometric mean only when the numbers are equal, since \(\tfrac{x + y}{2} - \sqrt{xy} = \tfrac{1}{2}(\sqrt{x} - \sqrt{y})^{2}\). So \(x = y\), and

\[ \frac{x}{y} + \frac{y}{x} = 1 + 1 = 2 \]

In fact \(x/y + y/x \geq 2\) for all positive \(x, y\), so 1 and 1/2 are impossible, and 4 would need \(x/y = 2 \pm \sqrt{3}\), which breaks the condition.

Answer: A

The minimum and maximum are -3 and 3.

\(\log|a| + \log|b| + \log|c| = \log|abc| = 0\), so \(|abc| = 1\). For integers this forces |a| = |b| = |c| = 1, so each of a, b and c is +1 or -1.

The sum is smallest when all three are -1, giving -3, and largest when all three are +1, giving 3. The other options leave out these extremes: -1 and 1 are sums with mixed signs, and 1 and 3 ignores that the values can be negative, since the logarithm is taken of the absolute value.

Answer: A

The value is \(2^{16}\).

Change the second logarithm to base 2. Since \(\sqrt{2} = 2^{1/2}\), \(\log_{\sqrt{2}} x = \frac{\log_{2} x}{1/2} = 2\log_{2} x\). So

\[ \begin{aligned} \log_{2} x + 2\log_{2} x &= 48 \\ 3\log_{2} x &= 48 \\ \log_{2} x &= 16,\quad x = 2^{16} \end{aligned} \]

\(4^{16} = 2^{32}\) doubles the exponent, and \(2^{14}\) and \(4^{14} = 2^{28}\) come from mishandling the base \(\sqrt{2}\), for example treating \(\log_{\sqrt{2}}\) as a factor other than 2.

Answer: C

The value of x is 256/81.

Convert the logarithm to its exponential form:

\[ \begin{aligned} \log_x \tfrac{9}{16} &= -\tfrac{1}{2} \\ x^{-1/2} &= \tfrac{9}{16} \\ x^{1/2} &= \tfrac{16}{9} \\ x &= \left(\tfrac{16}{9}\right)^{2} = \tfrac{256}{81} \end{aligned} \]

The base of a logarithm must be positive, so \(-3/4\) and \(-256/81\) are impossible. 4/3 is the square root of the answer, what you get by stopping one step early with the square root instead of squaring.

Answer: A

The value is 5.

Simplify inside the logarithm using powers of 5: \(125 = 5^3\), \(625 = 5^4\), \(25 = 5^2\).

\[ \begin{aligned} \frac{125 \times 625}{25} &= \frac{5^{3} \times 5^{4}}{5^{2}} = 5^{5} \\ \log_5 5^{5} &= 5 \end{aligned} \]

3125 is \(5^5\) itself, the number inside the logarithm rather than its logarithm. 6 and 25 come from adding or multiplying the exponents wrongly.

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.