Properties of Logarithms: Additional Practice (Set 2)

A cube root is the exponent $\tfrac{1}{3}$, so the power rule pulls it to the front first.

$\log_b \sqrt[3]{\frac{9}{16}} = \frac{1}{3}\log_b\!\left(\frac{9}{16}\right) = \frac{1}{3}\left(\log_b 9 - \log_b 16\right)$

Now write $9 = 3^2$ and $16 = 2^4$ and bring those exponents down as well.

$= \frac{1}{3}\left(2\log_b 3 - 4\log_b 2\right) = \frac{2q - 4p}{3}$

Combine the two logarithms with the quotient rule and simplify the fraction inside first.

$\log_2\!\left(4x^3\right) - \log_2(2x) = \log_2\!\left(\frac{4x^3}{2x}\right) = \log_2\!\left(2x^2\right)$

Now expand what is left, using $\log_2 2 = 1$ and the power rule.

$\log_2\!\left(2x^2\right) = \log_2 2 + 2\log_2 x = 1 + 2\log_2 x$

Take the base-10 logarithm and let the power rule drag the exponent down.

$\log_{10}\!\left(2^{50}\right) = 50\log_{10} 2 \approx 50(0.30103) = 15.0515$

That value lies between $15$ and $16$, so the number itself lies between $10^{15}$ and $10^{16}$.

$10^{15} < 2^{50} < 10^{16}$

A positive integer that is at least $10^{15}$ but less than $10^{16}$ has exactly $16$ digits. (Indeed $2^{50} = 1{,}125{,}899{,}906{,}842{,}624$.)

The exponent $50$ is not the answer, and neither is the logarithm $15.05$ on its own: the digit count is one more than the whole-number part.

You are not given $\log_b 3$ directly, but $6 = 2 \cdot 3$, so the quotient rule recovers it.

$\log_b 3 = \log_b 6 - \log_b 2 \approx 1.3 - 0.5 = 0.8$

Now $27 = 3^3$, so the power rule finishes the job.

$\log_b 27 = 3\log_b 3 \approx 3(0.8) = 2.4$

Tripling the given $1.3$ would give $3.9$, but that is $\log_b 216$, not $\log_b 27$: the cube belongs to the $3$, not to the $6$.

Write the whole argument as a single power of $x$: a cube root is the exponent $\tfrac{1}{3}$, and the reciprocal makes it negative.

$\frac{1}{\sqrt[3]{x}} = x^{-1/3}$

Now one application of the power rule finishes it.

$\log_b\!\left(x^{-1/3}\right) = -\frac{1}{3}\log_b x$

The reciprocal supplies the minus sign and the root supplies the $\tfrac{1}{3}$; both survive into the answer.

Change the base $\tfrac{1}{b}$ into $b$, where the value is known.

$\log_{1/b} x = \frac{\log_b x}{\log_b\!\left(\tfrac{1}{b}\right)}$

The denominator is the reciprocal case of the quotient rule: $\log_b\!\left(\tfrac{1}{b}\right) = -\log_b b = -1$.

$= \frac{2}{-1} = -2$

That is exactly right: $\left(\tfrac{1}{b}\right)^{-2} = b^2 = x$. Flipping the base flips the sign of every logarithm.

The product rule needs each factor positive, because the right side contains $\log_b M$ and $\log_b N$ separately, and each of those demands a positive argument.

A positive product is not enough. Take $M = N = -2$, so that $MN = 4 > 0$.

$\log_b(MN) = \log_b 4 \quad \text{exists, but} \quad \log_b(-2) \quad \text{does not.}$

The third statement fails on $b = 2$, $M = N = 4$, where $\log_2 8 = 3$ but $\log_2 4 + \log_2 4 = 4$. The fourth confuses a squared logarithm with a squared argument; it happens to hold only for the special values where $\log_b M$ is $0$ or $2$.

The quotient rule splits the fraction, and the whole denominator is subtracted.

$\log_b\!\left(\frac{\sqrt{x}}{y^2 z}\right) = \log_b \sqrt{x} - \log_b\!\left(y^2 z\right)$

The denominator is itself a product, so it splits again, and both of its logarithms inherit the minus sign.

$= \frac{1}{2}\log_b x - \left(2\log_b y + \log_b z\right) = \frac{1}{2}\log_b x - 2\log_b y - \log_b z$

Forgetting to distribute the minus sign over both factors of the denominator is the usual slip.

Condense the right side into a single logarithm, multiplying for the sum and dividing for the difference.

$\log_2 6 + \log_2 10 - \log_2 15 = \log_2\!\left(\frac{6 \cdot 10}{15}\right) = \log_2 4 = 2$

So $\log_2 N = 2$, and the definition of a logarithm turns that straight back into an exponential statement.

$N = 2^2 = 4$

Subtract the two magnitudes and let the quotient rule combine them. The reference intensity $S$ cancels.

$M_A - M_B = \log_{10}\!\left(\frac{I_A}{S}\right) - \log_{10}\!\left(\frac{I_B}{S}\right) = \log_{10}\!\left(\frac{I_A}{I_B}\right)$

We are told $I_A = 100 I_B$, so the ratio inside is $100$.

$M_A - M_B = \log_{10} 100 = 2$

A hundredfold jump in intensity is only $2$ points of magnitude, which is precisely why the scale is logarithmic: it compresses enormous ratios into small numbers.

Clear the coefficient first: $2\log_b 3 = \log_b 9$.

Now combine all three, multiplying the added arguments and dividing by the subtracted one.

$\log_b 9 + \log_b x - \log_b\!\left(x^2\right) = \log_b\!\left(\frac{9x}{x^2}\right) = \log_b\!\left(\frac{9}{x}\right)$

The coefficient $2$ becomes an exponent on $3$, giving $9$, not a factor of $6$.

Change the base $b^2$ into the base $b$, where the value is known.

$\log_{b^2} x = \frac{\log_b x}{\log_b\!\left(b^2\right)} = \frac{3}{2}$

The denominator is $2$ because $\log_b\!\left(b^2\right) = 2\log_b b = 2$.

The answer makes sense: squaring the base means you need only half the exponent to reach the same number, since $\left(b^2\right)^{3/2} = b^3 = x$.