Checking the Rule with Numbers

Show that $\log_2(4 \cdot 8) = \log_2 4 + \log_2 8$.

Evaluate the left side: $4 \cdot 8 = 32$ and $2^5 = 32$ = $\log_2 32 = 5$

Evaluate each log on the right side: $2^2 = 4$ and $2^3 = 8$ = $\log_2 4 = 2$, $\log_2 8 = 3$

Add and compare: $2 + 3 = 5$, the same as the left side = $5 = 5$

Answer: Both sides equal $5$: multiplying $4$ by $8$ adds the exponents $2$ and $3$.

Expanding a Logarithm

Expand $\log_3(9x)$ as far as possible ($x > 0$).

Split the product into a sum: $\log_3(9x) = \log_3 9 + \log_3 x$ = $\log_3 9 + \log_3 x$

Evaluate the log you can: $3^2 = 9$, so $\log_3 9 = 2$ = $2$

Write the final form: Replace $\log_3 9$ by $2$ = $2 + \log_3 x$

Answer: $\log_3(9x) = 2 + \log_3 x$

Condensing a Sum of Logarithms

Write $\log_4 2 + \log_4 8$ as a single logarithm and evaluate it.

Check that the bases match: Both logs have base $4$, so the rule applies = Same base

Multiply the arguments: $\log_4 2 + \log_4 8 = \log_4(2 \cdot 8)$ = $\log_4 16$

Evaluate: $4^2 = 16$ = $\log_4 16 = 2$

Answer: $\log_4 2 + \log_4 8 = \log_4 16 = 2$

Evaluating with Known Logs

Use $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$ (common logarithms, base 10) to approximate $\log 12$.

Factor the argument into numbers whose logs you know: $12 = 2 \cdot 2 \cdot 3$ = $\log 12 = \log 2 + \log 2 + \log 3$

Substitute the known values: $0.301 + 0.301 + 0.477$ = $1.079$

Check with a calculator: $\log 12 = 1.07918\ldots$ = $\log 12 \approx 1.079$

Answer: $\log 12 \approx 1.079$

Condensing with Variables

Write $\log_2 x + \log_2(x + 2)$ as a single logarithm, then evaluate it for $x = 2$.

Multiply the arguments: $\log_2 x + \log_2(x+2) = \log_2\big(x(x+2)\big)$ = $\log_2(x^2 + 2x)$

Note where it is defined: Both original logs need $x > 0$ and $x + 2 > 0$ = $x > 0$

Substitute $x = 2$: $2^2 + 2 \cdot 2 = 8$ = $\log_2 8$

Evaluate and check: $\log_2 8 = 3$, and $\log_2 2 + \log_2 4 = 1 + 2 = 3$ = $3$

Answer: $\log_2(x^2 + 2x)$ for $x > 0$; at $x = 2$ its value is $3$.

Mistake: Writing $\log(M + N) = \log M + \log N$

Why: The rule is about a product inside the log, not a sum. Check: $\log(6 + 4) = \log 10 = 1$, but $\log 6 + \log 4 \approx 0.778 + 0.602 = 1.380$.

Correct: Only $\log(MN) = \log M + \log N$. A log of a sum, like $\log(x + 3)$, cannot be split.

Mistake: Multiplying the logs: $\log_b(MN) = \log_b M \cdot \log_b N$

Why: Multiplying the numbers adds the exponents, it does not multiply them. Check: $\log_2 4 \cdot \log_2 8 = 2 \cdot 3 = 6$, but $\log_2 32 = 5$.

Correct: A product inside becomes a sum outside: $\log_2(4 \cdot 8) = 2 + 3 = 5$.

Mistake: Combining logs with different bases: $\log_2 x + \log_3 y = \log_6(xy)$

Why: The proof uses $b^m \cdot b^n = b^{m+n}$, which only works when both powers have the same base $b$.

Correct: Condense only logs with the same base. $\log_2 x + \log_3 y$ stays as it is (or use the Change of Base Formula first).

Mistake: Using the rule with negative factors: $\log\big((-2)(-5)\big) = \log(-2) + \log(-5)$

Why: $\log 10 = 1$ exists, but $\log(-2)$ and $\log(-5)$ do not: the argument of a logarithm must be positive.

Correct: The rule needs $M > 0$ and $N > 0$. When you condense, keep the domain of the original logs (for example $x > 0$).

Decibels and Sound Intensity

Sound level in decibels is $L = 10 \log\left(\frac{I}{I_0}\right)$. When the intensity $I$ is multiplied by a factor $k$, the product rule gives $\log(k \cdot \frac{I}{I_0}) = \log k + \log\frac{I}{I_0}$, so the level goes up by $10 \log k$.

A sound that becomes 10 times more intense goes up by $10 \log 10 = 10$ decibels, whatever level it started at.

Slide Rules: Multiplying by Adding

A slide rule marks each number $x$ at a distance $\log x$ from the start. Sliding one scale along the other adds two distances, and by the product rule the sum of the distances is the log of the product.

To multiply $2 \cdot 3$, add $\log 2 + \log 3 \approx 0.301 + 0.477 = 0.778$, which is where $6$ is marked.

Bits of Information

In computer science, the number of bits needed to store one of $N$ equally likely options is $\log_2 N$. Combining two independent choices multiplies the options but adds the bits.

A choice from $8$ symbols needs $\log_2 8 = 3$ bits, a choice from $4$ needs $2$ bits, and both together have $32$ options: $\log_2 32 = 3 + 2 = 5$ bits.

Q: Does the product rule work for more than two factors?

A: Yes. Apply it again and again: $\log_b(MNP) = \log_b M + \log_b N + \log_b P$. For example $\log 42 = \log 2 + \log 3 + \log 7$.

Q: Does it work for the natural logarithm too?

A: Yes, it works for every valid base, including $e$: $\ln(MN) = \ln M + \ln N$. For example $\ln 2 + \ln 5 = \ln 10$.

Q: Is there a rule for $\log(M + N)$?

A: No. A logarithm of a sum has no simpler form in general. $\log(6 + 4) = \log 10 = 1$, while $\log 6 + \log 4 \approx 1.380$.

Q: How is this different from the quotient and power rules?

A: They are siblings of the same idea. The product rule turns multiplication into addition; the Quotient Rule for Logarithms turns division into subtraction; the Power Rule for Logarithms turns a power into multiplication by the exponent.