Laws of natural logs and how to use them

The natural log follows the same rules as any logarithm. Since its base is e, it also helps us undo an exponential with base e. Let’s use those two ideas to combine a few expressions and keep track of which inputs work.

Use the natural log to undo an exponential

The natural log is $\ln x = \log_{e} x$ . If $e^{x} = y$ , then $\ln y = x$ . Since $e^{x}$ is always positive, $\ln (e^{x}) = x$ for every real $x$ . Going in the other direction, $e^{\ln x} = x$ for $x > 0$ .

Product rule for natural logs

For $x > 0$ and $y > 0$ , $\ln (x y) = \ln x + \ln y$ . We can use the rule in either direction. Here we use it to write two natural logs as one.

\begin{array}{l} \ln (64) + \ln (16) = \ln (64 \cdot 16) \\ = \ln (1024) \approx 6.9315 \end{array}

Quotient rule for natural logs

For $x > 0$ and $y > 0$ , $\ln (x / y) = \ln x - \ln y$ . The first log supplies the numerator and the second supplies the denominator.

\begin{array}{l} \ln (32) - \ln (8) = \ln (\frac{32}{8}) \\ = \ln (4) \approx 1.3863 \end{array}

Power rule for natural logs

For $x > 0$ and real $k$ , $\ln (x^{k}) = k \ln x$ . Bring the powers in front to turn the difference into a subtraction of coefficients.

\begin{array}{l} \ln (8^{4}) - \ln (8^{2}) = 4 \ln (8) - 2 \ln (8) \\ = 2 \ln (8) = \ln (64) \end{array}

Since $64 = 2^{6}$ , we can also write that answer as $6 \ln (2)$ .

Keep the domain when an expression contains a square

A real logarithm requires a positive argument. In $\ln (x^{2})$ , the square is positive for every $x \neq 0$ . Since $x^{2} = | x |^{2}$ and $| x |$ is positive on that domain, we can apply the power rule to $| x |$ .

\ln (x^{2}) = 2 \ln | x | x \neq 0

For example, $x = - 3$ gives $\ln (9) = 2 \ln (3)$ . For the product or quotient rule, each argument after splitting a log also has to be positive.

Read the domain and special values from the graph

The graph crosses the x-axis at (1, 0). Its vertical asymptote is x = 0.

The graph is defined for $x > 0$ and crosses at $x = 1$ because $\ln 1 = 0$ . As $x$ approaches zero from the right, $\ln x$ falls without bound. As $x$ grows, $\ln x$ increases without bound.

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