Evaluating with the Power Rule

Evaluate $\log_2(8^5)$ without a calculator.

Move the exponent to the front: $\log_2(8^5) = 5 \cdot \log_2 8$ = $5 \log_2 8$

Evaluate the simpler logarithm: $2^3 = 8$, so $\log_2 8 = 3$ = $\log_2 8 = 3$

Multiply: $5 \cdot 3 = 15$ = $15$

Answer: $\log_2(8^5) = 15$. Check: $8^5 = (2^3)^5 = 2^{15}$.

Logarithm of a Root

Evaluate $\log_5 \sqrt{125}$.

Write the root as a fractional power: $\sqrt{125} = 125^{1/2}$ = $\log_5(125^{1/2})$

Apply the power rule: $\log_5(125^{1/2}) = \frac{1}{2} \log_5 125$ = $\frac{1}{2} \log_5 125$

Evaluate the logarithm: $5^3 = 125$, so $\log_5 125 = 3$ = $\log_5 125 = 3$

Multiply: $\frac{1}{2} \cdot 3 = \frac{3}{2}$ = $\frac{3}{2}$

Answer: $\log_5 \sqrt{125} = \frac{3}{2} = 1.5$

Expanding with All Three Rules

Expand $\log \dfrac{x^3 y^2}{\sqrt{z}}$ (all variables positive).

Quotient rule: numerator minus denominator: $\log(x^3 y^2) - \log \sqrt{z}$ = Two logarithms

Product rule on the numerator: $\log(x^3 y^2) = \log x^3 + \log y^2$ = $\log x^3 + \log y^2 - \log \sqrt{z}$

Write the root as a power: $\sqrt{z} = z^{1/2}$ = $\log x^3 + \log y^2 - \log z^{1/2}$

Power rule on every term: Bring each exponent to the front = $3 \log x + 2 \log y - \frac{1}{2} \log z$

Answer: $\log \dfrac{x^3 y^2}{\sqrt{z}} = 3 \log x + 2 \log y - \frac{1}{2} \log z$

Condensing into One Logarithm

Write $2 \ln x + 3 \ln y - 4 \ln z$ as a single logarithm.

Power rule backwards: coefficients become exponents: $2 \ln x = \ln x^2$, $3 \ln y = \ln y^3$, $4 \ln z = \ln z^4$ = $\ln x^2 + \ln y^3 - \ln z^4$

Product rule backwards on the sum: $\ln x^2 + \ln y^3 = \ln(x^2 y^3)$ = $\ln(x^2 y^3) - \ln z^4$

Quotient rule backwards on the difference: $\ln(x^2 y^3) - \ln z^4 = \ln \dfrac{x^2 y^3}{z^4}$ = $\ln \dfrac{x^2 y^3}{z^4}$

Answer: $2 \ln x + 3 \ln y - 4 \ln z = \ln \dfrac{x^2 y^3}{z^4}$

Approximating with Known Logarithms

Use $\log 2 \approx 0.301$ and $\log 3 \approx 0.477$ to approximate $\log 72$.

Factor into prime powers: $72 = 8 \cdot 9 = 2^3 \cdot 3^2$ = $\log 72 = \log(2^3 \cdot 3^2)$

Product rule: $\log(2^3 \cdot 3^2) = \log 2^3 + \log 3^2$ = $\log 2^3 + \log 3^2$

Power rule: $\log 2^3 + \log 3^2 = 3 \log 2 + 2 \log 3$ = $3 \log 2 + 2 \log 3$

Substitute the values: $3 \cdot 0.301 + 2 \cdot 0.477 = 0.903 + 0.954$ = $1.857$

Answer: $\log 72 \approx 1.857$ (a calculator gives $1.8573$)

Mistake: Treating $(\log M)^p$ as $p \log M$

Why: The power rule only works when the exponent is on the **argument**, inside the logarithm. In $(\log M)^p$ the whole logarithm is raised to a power. Example: $(\log 100)^3 = 2^3 = 8$, but $3 \log 100 = 6$.

Correct: $\log(M^p) = p \log M$, while $(\log M)^p$ cannot be simplified with the power rule. Look where the exponent sits before you move it.

Mistake: Applying the exponent to the whole argument: $\log(3x^2) = 2 \log(3x)$

Why: In $3x^2$ only $x$ is squared, not $3$. Check with $x = 10$: $\log 300 \approx 2.48$, but $2 \log 30 \approx 2.95$.

Correct: Split the product first: $\log(3x^2) = \log 3 + \log x^2 = \log 3 + 2 \log x$.

Mistake: Turning a square root into a factor of 2: $\log \sqrt{x} = 2 \log x$

Why: A square root is the power $\frac{1}{2}$, not the power $2$. The index of the root goes into the denominator.

Correct: $\log \sqrt{x} = \log x^{1/2} = \frac{1}{2} \log x$, and $\log \sqrt[3]{x} = \frac{1}{3} \log x$.

Mistake: Condensing $3 \log x - \log y$ into $\log \frac{3x}{y}$

Why: A coefficient in front of a logarithm is an **exponent** of the argument, not a factor of it.

Correct: First move the coefficient up: $3 \log x = \log x^3$. Then $3 \log x - \log y = \log \frac{x^3}{y}$.

Loudness in Decibels

The change in sound level is $\Delta L = 10 \log \frac{I_2}{I_1}$ decibels, where $\frac{I_2}{I_1}$ is how many times the intensity grows. The power rule turns powers of 10 into simple multiplication.

Doubling your distance from a speaker divides the intensity by $2^2 = 4$, so $\Delta L = 10 \log(2^{-2}) = -20 \log 2 \approx -6$ dB.

Counting Digits of Huge Numbers

A whole number $N$ has $\lfloor \log N \rfloor + 1$ digits. With the power rule you can count the digits of enormous powers without computing them.

$\log(2^{10}) = 10 \log 2 \approx 3.01$, so $2^{10} = 1024$ has $3 + 1 = 4$ digits.

Population Growth

A population that grows by a fixed percentage every year is an exponential function of time. To find when it reaches a target, the power rule brings the time down from the exponent.

At 10% growth per year, a population doubles when $1.1^t = 2$: $t \log 1.1 = \log 2$, so $t = \frac{\log 2}{\log 1.1} \approx 7.3$ years.

Q: Why does the power rule work?

A: A logarithm is an exponent. If $\log_b M = x$, then $M = b^x$ and $M^p = b^{px}$. Raising to the power $p$ multiplies the exponent by $p$, so the logarithm (the exponent) is multiplied by $p$ too.

Q: Does the exponent have to be a whole number?

A: No. The power rule works for any real exponent: negative ($\log_b \frac{1}{M^2} = -2 \log_b M$), fractional ($\log_b \sqrt[3]{M} = \frac{1}{3} \log_b M$) and even irrational ones.

Q: Are $\log(x^2)$ and $2 \log x$ really the same function?

A: Only for $x > 0$. The function $\log(x^2)$ also exists for negative $x$ (for example $\log((-10)^2) = \log 100 = 2$), but $2 \log x$ does not. For every $x \neq 0$ the exact statement is $\log(x^2) = 2 \log |x|$. The graph above shows both curves.

Q: When should I expand and when should I condense?

A: Expand when you want simple pieces, for example to approximate $\log 72$ from $\log 2$ and $\log 3$. Condense when you want one logarithm, for example to solve an equation like $2 \log x - \log 4 = 2$ (see Solving Logarithmic Equations).