6.3 Logarithms and the laws of logarithms — Learn

Prerequisites: the graphs of $a^x$ and $e^x$ from Section 6.1; the laws of indices from Chapter 2; the idea of an inverse function from Chapter 2.

By the end of this section you can: convert between exponential and logarithmic form; evaluate a logarithm by writing the input as a power; sketch $y = \ln x$ ; use the three laws of logarithms in both directions; and change the base of a logarithm.

1. A logarithm undoes a power

A logarithm answers the question: what power of the base gives this number?

\log_a x = y \quad \text{means} \quad a^y = x.

The base $a$ must be positive and not equal to $1$ . The input $x$ must be positive, because $a^y$ is always positive. This is the domain of a logarithm: $x > 0$ .

Example 1. Write $2^5 = 32$ in logarithmic form.

The base $2$ , the power $5$ and the result $32$ give

\log_2 32 = 5.

Example 2. Work out $\log_3 81$ .

Ask what power of $3$ gives $81$ . Since $81 = 3^4$ , the answer is $4$ .

Example 3. Work out $\log_2 \dfrac{1}{8}$ .

Write the input as a power of the base: $\dfrac{1}{8} = 2^{-3}$ . So $\log_2 \dfrac{1}{8} = -3$ . A logarithm of a number smaller than $1$ is negative.

2. The graph of a logarithm

The logarithmic function $y = \log_a x$ is the inverse of $y = a^x$ . Its graph is the reflection of the exponential graph in the line $y = x$ . So:

it passes through $(1, 0)$ , because $\log_a 1 = 0$ ;
it passes through $(a, 1)$ , because $\log_a a = 1$ ;
the $y$ -axis is a vertical asymptote, because the input must stay positive;
the domain is $x > 0$ and the range is all real numbers;
when $a > 1$ the graph rises, and when $0 < a < 1$ it falls.

The natural logarithm, written $\ln x$ , is the logarithm to base $e$ :

\ln x = \log_e x.

Its graph is the reflection of $y = e^x$ in the line $y = x$ . Because $e$ is the natural base for growth, $\ln x$ is the natural inverse.

3. The laws of logarithms

The three laws follow from the laws of indices. For positive $x$ and $y$ :

\log_a(xy) = \log_a x + \log_a y,

\log_a\dfrac{x}{y} = \log_a x - \log_a y,

\log_a\left(x^k\right) = k\log_a x.

The first law says that adding logarithms multiplies the inputs. The second says that subtracting logarithms divides the inputs. The third says that a multiplier becomes a power. The power law holds for every real $k$ , including $k = -1$ and $k = -\dfrac{1}{2}$ .

Two useful special values follow from the definition:

\log_a a = 1, \qquad \log_a 1 = 0.

Example 4. Write $\log_2 8 + \log_2 4$ as a single logarithm, and evaluate it.

By the product law, $\log_2 8 + \log_2 4 = \log_2 32 = 5$ . Checking directly: $\log_2 8 = 3$ and $\log_2 4 = 2$ , and $3 + 2 = 5$ .

Example 5. Expand $\ln\dfrac{x^3}{y}$ using the laws of logarithms.

The quotient law splits the fraction, and the power law brings the $3$ down:

\ln\frac{x^3}{y} = \ln x^3 - \ln y = 3\ln x - \ln y.

Example 6. Write $2\ln x + \ln y$ as a single logarithm.

The power law turns the first term into $\ln x^2$ . The product law then combines the two terms:

2\ln x + \ln y = \ln x^2 + \ln y = \ln\left(x^2y\right).

Example 7. Solve $\log_5 x = 2$ .

Rewrite the logarithm as a power: $x = 5^2 = 25$ . The answer is positive, as the domain requires.

Example 8. Change the base of $\log_2 7$ to base $e$ .

Write $y = \log_2 7$ , so $2^y = 7$ . Take the natural logarithm of both sides: $y\ln 2 = \ln 7$ , giving

\log_2 7 = \frac{\ln 7}{\ln 2}.

This change-of-base rule turns any base into a base your calculator can handle. Section 6.4 uses it to solve $a^x = b$ .

Watch out. These mistakes are common.

Treating a logarithm as a multiplier. The symbol $\log$ is a function, not a number. So $\ln x$ is not $\ln \times x$ , and $\ln 2x$ is not $2\ln x$ .

Adding inside instead of multiplying. The product law is $\log(xy) = \log x + \log y$ . There is no simple law for $\log(x + y)$ ; in general $\log(x + y)$ is not $\log x + \log y$ . Test it with $x = y = 10$ : $\log_{10}20 \approx 1.30$ , not $2$ .

Ignoring the domain. A logarithm needs a positive input. The equation $\log_2 x = 3$ has the single solution $x = 8$ , because a negative or zero input is not allowed.

Re-check: after using a law, choose a value for the letters and check both sides with a calculator. For example, test $\ln 6 = \ln 2 + \ln 3$ .

You should now be able to

convert between $a^y = x$ and $\log_a x = y$ ;
evaluate a logarithm by writing the input as a power of the base;
sketch $y = \ln x$ and state its intercept, asymptote and domain;
use the product, quotient and power laws in both directions;
state the domain of a logarithmic expression;
change the base of a logarithm.