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Logarithm rules

There are three logarithm rules, and none of them is a new fact. Each one is an exponent law read through the definition $\log_b M = m$ means $b^m = M$, which is the sentence the whole subject rests on: a logarithm is an exponent.

$$\log_b(MN) = \log_b M + \log_b N$$ $$\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N$$ $$\log_b(M^p) = p\log_b M$$

Throughout, $b > 0$ and $b \neq 1$, and $M$ and $N$ are strictly positive. That last condition is where solutions get lost, and it is worth carrying through every step rather than remembering at the end.

Deriving the three, not listing them

Write $m = \log_b M$ and $n = \log_b N$. By definition that means $M = b^m$ and $N = b^n$. Every derivation below is the same move: turn logs into exponents, use an exponent law, turn back.

Product. Multiply the two:

$$MN = b^m \cdot b^n = b^{m+n}$$

The middle step is the exponent law $b^m b^n = b^{m+n}$, which is true because multiplying $m$ copies of $b$ by $n$ copies gives $m + n$ copies. Reading $MN = b^{m+n}$ back through the definition, the exponent that produces $MN$ is $m + n$, so $\log_b(MN) = m + n = \log_b M + \log_b N$.

Quotient. Divide instead:

$$\frac{M}{N} = \frac{b^m}{b^n} = b^{m-n}$$

so $\log_b(M/N) = m - n$. The special case $M = 1$ gives $\log_b(1/N) = -\log_b N$, since $\log_b 1 = 0$.

Power. Raise $M$ to a power $p$:

$$M^p = (b^m)^p = b^{mp}$$

using $(b^m)^p = b^{mp}$, so $\log_b(M^p) = mp = p\log_b M$. Roots are included free, because $\sqrt[n]{M} = M^{1/n}$, so $\log_b \sqrt{M} = \tfrac{1}{2}\log_b M$.

That is the entire content. The product rule turns multiplication into addition, the power rule turns exponentiation into multiplication, and each operation drops one level in difficulty - which is what logarithms were invented to do.

Change of base

A calculator offers $\log_{10}$ and $\ln$, and a problem asks for $\log_4 27$. Let $y = \log_4 27$, so $4^y = 27$. Take the natural log of both sides and use the power rule on the left:

$$y\ln 4 = \ln 27 \qquad\Longrightarrow\qquad y = \frac{\ln 27}{\ln 4}$$

Nothing about base $e$ mattered, so the general statement holds for any legal base $c$:

$$\log_b M = \frac{\log_c M}{\log_c b}$$

Numerically, $\ln 27 = 3.295837$ and $\ln 4 = 1.386294$, so $\log_4 27 = 3.295837/1.386294 = 2.377444$.

Check it a second way, by a route that never touches $\ln$. Since $27 = 3^3$, the power rule gives $\log_4 27 = 3\log_4 3$, and change of base into base 2 gives $\log_4 3 = \log_2 3 / \log_2 4 = 1.58496250/2 = 0.79248125$. Then $3 \times 0.79248125 = 2.37744375$, which is $2.377444$ to six figures. Same number, by different arithmetic. A third check is available for nothing: $4^{2.377444} = 27.000$.

Change of base practice runs the same conversion in both directions.

Expanding and condensing

Textbooks file these three under the properties of logarithms, and the two directions they get used in have names. Expanding means running the log rules left to right until every logarithm has a single variable inside it:

$$\log_b\!\left(\frac{x^3\sqrt{y}}{z^2}\right) = 3\log_b x + \tfrac{1}{2}\log_b y - 2\log_b z$$

The quotient rule handles the fraction bar, the product rule splits the numerator, and the power rule pulls each exponent out front. Condensing is the same sequence run backwards, and it is what turns an equation with several logarithms into an equation with one.

Solving an exponential equation uses the power rule and nothing else. Take $4^{3x-1} = 27$. The unknown is in the exponent, so take a logarithm of both sides to bring it down:

$$(3x - 1)\log 4 = \log 27 \qquad\Longrightarrow\qquad 3x - 1 = \frac{\log 27}{\log 4} = 2.3774438$$

That is the number computed above, so $3x = 3.3774438$ and $x = 1.1258146$. Substituting back, $3(1.1258146) - 1 = 2.3774438$ and $4^{2.3774438} = 27.000$. Practice at solving exponential equations varies which side the unknown sits on; the move is always the same one.

The two errors that actually cost marks

There is no rule for $\log(a + b)$. The product rule turns a product into a sum of logs; it says nothing about a sum. Write $\log(a + b) = \log a + \log b$ and the numbers disagree immediately. Take $a = 2$, $b = 8$ in base 10. The left side is $\log_{10}(10) = 1$ exactly. The right side is $\log_{10} 2 + \log_{10} 8 = \log_{10} 16 = 1.204120$. The gap is $\log_{10}(16/10) = \log_{10}1.6$, and in general the difference between the two sides is $\log(ab) - \log(a+b)$, which is almost never zero. The same warning covers two relatives: $\dfrac{\log a}{\log b}$ is not $\log(a/b)$ - it is $\log_b a$ - and $(\log x)^2$ is not $2\log x$.

Condensing widens the domain, so check every candidate. Solve $\log_2 x + \log_2(x - 3) = 2$. Condense the left side with the product rule:

$$\log_2\!\big(x(x-3)\big) = 2 \quad\Longrightarrow\quad x^2 - 3x = 2^2 = 4$$

$$x^2 - 3x - 4 = 0 \quad\Longrightarrow\quad (x - 4)(x + 1) = 0$$

so $x = 4$ or $x = -1$. Test them in the original equation, not the condensed one. At $x = 4$: $\log_2 4 + \log_2 1 = 2 + 0 = 2$, correct. At $x = -1$: $\log_2(-1)$ does not exist, and neither does $\log_2(-4)$. Reject it.

The reason is exact and worth seeing rather than treating as a ritual. The original left side needs $x > 0$ and $x - 3 > 0$, so $x > 3$. The condensed version only needs $x(x-3) > 0$, which is satisfied by $x > 3$ or by $x < 0$. Condensing quietly admitted the whole negative half-line, and the quadratic handed back a root from the region that was never legal. Every extraneous solution in a logarithmic equation arrives this way, so the check is not optional caution - it is the step that undoes the domain change you made.

Expanding, condensing and the three rules together are a single skill, and practising them as one is more useful than drilling them separately. Our study guide for OpenStax Calculus 1.5 picks the same properties up where calculus needs them, in logarithmic differentiation and in rewriting $a^x$ as $e^{x\ln a}$.

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