Properties of Logarithms: Free Response

Start from what $t$ means: $b^t = M$. Raise both sides to the power $p$; equal numbers raised to the same power stay equal, so

$M^p = \left(b^t\right)^p.$

The given exponent law collapses the right side into a single power of $b$:

$M^p = b^{\,tp}.$

Since $b > 0$, the quantity $b^{tp}$ is positive, so $M^p$ is positive and its logarithm exists. Reading the last line back through the definition of a logarithm says the exponent that produces $M^p$ is $tp$, so $\log_b\!\left(M^p\right) = tp$. Substituting $t = \log_b M$ finishes it:

$\log_b\!\left(M^p\right) = p\log_b M.$

Nothing in the argument used a specific value of $b$, $M$, or $p$, so the identity holds for every valid base, every positive $M$, and every real exponent $p$.

$121 = 11^2$, so $\log_{11}121 = 2$ directly from the definition, since $11^2 = 121$. The power rule pulls the outer exponent to the front:

$\log_{11}\!\left(121^3\right) = 3\log_{11}121 = 3(2) = 6.$

Check it against the definition: $11^6 = \left(11^2\right)^3 = 121^3$, so $6$ really is the exponent that produces $121^3$.

Substitute $x = -9$ into both sides.

Left side: $x^2 = (-9)^2 = 81$, a positive number, so $\log_b(81)$ is a perfectly good real number.

Right side: $2\log_b x = 2\log_b(-9)$, and $-9$ is not in the domain of a logarithm, which only accepts positive inputs. So

$x^2 = 81 \ (\text{defined}), \qquad \log_b(-9) \ (\text{undefined}),$

and the right side is not a number at all.

Trace this back to part A: the proof opens by setting $t = \log_b M$ (here $M = x = -9$), a step that only makes sense when $M > 0$. That is exactly the step that fails, not some later piece of algebra. The identity that survives for every $x \neq 0$ replaces the bare $\log_b x$ with $\log_b|x|$: $\log_b\!\left(x^2\right) = 2\log_b|x|$.

Split the fraction first with the quotient rule:

$\log_4\!\left(\frac{16a^3}{\sqrt[3]{c}}\right) = \log_4\!\left(16a^3\right) - \log_4\sqrt[3]{c}.$

The numerator is a product, so the product rule splits it again, and the cube root is the power $c^{1/3}$:

$= \log_4 16 + \log_4\!\left(a^3\right) - \log_4\!\left(c^{1/3}\right).$

Now the power rule brings both exponents to the front, and $\log_4 16 = 2$ since $4^2 = 16$:

$\log_4\!\left(\frac{16a^3}{\sqrt[3]{c}}\right) = 2 + 3\log_4 a - \frac{1}{3}\log_4 c.$

The product and quotient rules only combine logarithms with a coefficient of $1$, so clear every coefficient first by running the power rule backward:

$2\log_b p - \frac{1}{3}\log_b q + \log_b r = \log_b\!\left(p^2\right) - \log_b\!\left(q^{1/3}\right) + \log_b r.$

The two logarithms being added combine into a product, and the one being subtracted goes into the denominator:

$= \log_b\!\left(\frac{p^2 \cdot r}{q^{1/3}}\right) = \log_b\!\left(\frac{p^2 r}{\sqrt[3]{q}}\right).$

The product rule reads $\log_b M + \log_b N = \log_b(MN)$ and the quotient rule reads $\log_b M - \log_b N = \log_b(M/N)$. Both patterns need a bare logarithm, with no coefficient in front, on each side. A term like $-\tfrac{1}{3}\log_b q$ does not match that pattern, so it cannot be folded in immediately.

If $2\log_b p$ and $\log_b r$ are condensed first, the result $\log_b\!\left(p^2 r\right)$ is correct as far as it goes, but the leftover term still carries its coefficient:

$\log_b\!\left(p^2 r\right) - \frac{1}{3}\log_b q.$

This is not a dead end: the power rule can still clear that one coefficient, $-\tfrac{1}{3}\log_b q = -\log_b\!\left(q^{1/3}\right)$, and the quotient rule then finishes the job, reaching the same single logarithm either way. The real cost of combining out of order is one extra step, not an impossibility. Clearing every coefficient up front avoids that extra step: once every term is already a bare logarithm, the product and quotient rules apply cleanly to all of them in a single pass.

Change of base puts the argument on top and the target base underneath:

$\log_2 n = \frac{\log_{10} n}{\log_{10} 2}.$

Substitute $n = 1{,}000{,}000$, using $\log_{10}\!\left(1{,}000{,}000\right) = 6$ exactly and $\log_{10}2 \approx 0.3010$:

$\log_2\!\left(1{,}000{,}000\right) = \frac{6}{0.3010} \approx 19.93.$

A worst-case search needs about $20$ comparisons for a million items.

Write the new list size as the old size times the growth factor $8$, and apply the product rule:

$\log_2(8n) = \log_2 8 + \log_2 n = 3 + \log_2 n,$

since $2^3 = 8$ gives $\log_2 8 = 3$. Comparing to the old comparison count $\log_2 n$, the new count is exactly $3$ more than the old one, whatever $n$ happened to be. Growing the list eightfold costs exactly $3$ extra comparisons, a number that does not depend on the starting size $n$.

Write the new list size as $kn$ and apply the product rule directly, with no restriction on the size of $n$ beyond $n > 0$:

$\log_2(kn) = \log_2 k + \log_2 n.$

Subtract the old comparison count $\log_2 n$ from the new one $\log_2(kn)$:

$\log_2(kn) - \log_2 n = \log_2 k.$

The right side depends only on the growth factor $k$; $n$ has cancelled completely, because it appeared on both sides in exactly the same way. For $k > 1$, $\log_2 k > 0$, so this really is an increase, of exactly $\log_2 k$ comparisons, regardless of how large the list already was. (The same algebra holds for $0 < k < 1$; there $\log_2 k < 0$, so shrinking the list by a factor $k$ decreases the comparison count instead, by the same fixed amount.) This is the same fact the lesson's logarithmic-scale picture shows geometrically: the distance added by multiplying by a fixed factor is the same distance wherever you start.

Compute the product first: $MN = (-4)(9) = -36$, a negative number, so $\log_b(MN) = \log_b(-36)$ does not exist; the left side already fails before the rule is even applied.

Check the right side term by term. $N = 9$ is positive, so $\log_b N = \log_b 9$ is a perfectly good real number. But $M = -4$ is negative, so $\log_b M = \log_b(-4)$ does not exist. A sum with one undefined term has no value, so

$\log_b M + \log_b N \text{ is undefined too},$

even though one of its two pieces, $\log_b 9$, is perfectly fine by itself.

Let $M < 0$ and $N > 0$ be any such pair. Their product is negative: a negative number times a positive number is always negative, so

$MN < 0$

for every such pair, which means $\log_b(MN)$ never exists, no matter which $M$ and $N$ you pick.

On the right side, $\log_b M$ requires $M > 0$, and $M < 0$ by assumption, so $\log_b M$ never exists, for any negative $M$ whatsoever, whatever $N$ happens to be. A sum is only defined when both of its terms are, so

$\log_b M + \log_b N \text{ never exists, for any } M < 0 \text{ and any } N > 0,$

even though $\log_b N$ on its own is always fine. Both sides of the product rule fail here, together, unlike the case where $M$ and $N$ are both negative, where the product $MN$ is positive and only the right side breaks.

The pair tested in parts A and B shows that once $M$ is negative, both sides of the product rule can fail: the left side $\log_b(MN)$ fails because the product itself turns negative, and the right side fails independently, because $\log_b M$ alone is already undefined. That is a different failure mode from the case where $M$ and $N$ are both negative, where the left side survives (a negative times a negative is positive) while only the right side breaks; dropping positivity can take down either side, or both, depending on how it is dropped.

What neither case does is refute the product rule as it was actually stated. The rule is a conditional claim, true whenever its hypothesis holds:

$M > 0 \text{ and } N > 0 \;\Longrightarrow\; \log_b(MN) = \log_b M + \log_b N.$

A pair that fails the hypothesis falls outside the claim entirely, whichever way it fails, and there is no promise there to break. Testing the rule honestly means checking positive $M$ and $N$, which is exactly where the proof, and every worked example in the lesson, lives.

Apply change of base with $a = b$:

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

Simplify the denominator with the reciprocal rule and $\log_b b = 1$:

$\log_b\!\left(\frac{1}{b}\right) = -\log_b b = -1.$

Substitute that back in:

$\log_{1/b} M = \frac{\log_b M}{-1} = -\log_b M.$

Nothing here used a specific $b$ or $M$: the identity holds for every valid base $b$ and every $M > 0$.

Apply the identity with $b = 5$ and $M = 125$:

$\log_{1/5}125 = -\log_5 125.$

Since $5^3 = 125$, $\log_5 125 = 3$, so

$\log_{1/5}125 = -3.$

Check it directly against the definition: $\left(\tfrac{1}{5}\right)^{-3} = 5^3 = 125$, confirming that $-3$ is exactly the exponent that turns $\tfrac{1}{5}$ into $125$.

A reflection across the $x$-axis sends every point $(x, y)$ to $(x, -y)$; algebraically, that means the reflected function's value at any input is the negative of the original function's value there.

The identity from part A says exactly that: for every input $M > 0$,

$\log_{1/b}M = -\log_b M,$

so the value of $y = \log_{1/b}x$ at any $x$ is the negative of the value of $y = \log_b x$ at that same $x$. That is the algebraic definition of a reflection across the $x$-axis, so the picture is not a separate fact to memorize; it is the identity from part A, read geometrically.