Logarithmic laws

Logarithmic laws are rules that tell you how to simplify and manipulate expressions involving logarithms. They are the logarithmic counterparts of the exponent laws. In this chapter, we will:

• State the main logarithm laws
• Show why each law makes sense using exponent rules
• Give simple examples (both symbolic and numerical)
• Point out common mistakes to avoid

Throughout, assume the base $b$ is positive and $b \ne 1$ (this is required for logarithms to be defined as functions).

The product law

Statement

For any positive $M$ and $N$ and base $b$,
$$
\log_b(MN) = \log_b M + \log_b N.
$$

So, the logarithm of a product is the sum of the logarithms.

Why it matches exponent rules

If $b^x = M$ and $b^y = N$, then
$$
MN = b^x \cdot b^y = b^{x+y}.
$$
Taking $\log_b$ of both sides,
$$
\log_b(MN) = \log_b(b^{x+y}) = x + y.
$$
But $x = \log_b M$ and $y = \log_b N$, so
$$
\log_b(MN) = \log_b M + \log_b N.
$$

Examples

• With base $10$:
  $$
  \log_{10}(1000) = \log_{10}(10 \cdot 100)
  = \log_{10} 10 + \log_{10} 100
  = 1 + 2 = 3.
  $$
• Symbolic:
  $$
  \log_3(9x) = \log_3 9 + \log_3 x.
  $$

Common mistakes

• Wrong: $\log_b(M+N) = \log_b M + \log_b N$.
  This is false. There is no simple law for $\log_b(M+N)$.
• You can only apply this when you have a product inside the log, not outside:
• $\log_b(kM)$ can become $\log_b k + \log_b M$ (product inside).
• But $k \log_b M$ is outside multiplication; it’s different and uses a different law (power law, below, in reverse).

The quotient law

Statement

For any positive $M$ and $N$ and base $b$,
$$
\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N.
$$

So, the logarithm of a quotient is the difference of the logarithms.

Why it matches exponent rules

If $b^x = M$ and $b^y = N$, then
$$
\frac{M}{N} = \frac{b^x}{b^y} = b^{x-y}.
$$
Taking $\log_b$ of both sides,
$$
\log_b\left(\frac{M}{N}\right) = \log_b(b^{x-y}) = x - y
= \log_b M - \log_b N.
$$

Examples

• With base $10$:
  $$
  \log_{10}\left(\frac{1000}{10}\right)
  = \log_{10} 1000 - \log_{10} 10
  = 3 - 1 = 2.
  $$
• Symbolic:
  $$
  \log_2\left(\frac{8x}{y}\right)
  = \log_2 8 + \log_2 x - \log_2 y.
  $$

Common mistakes

• Wrong: $\log_b\left(\frac{M}{N}\right) = \frac{\log_b M}{\log_b N}$.
  This is false. The correct expression is a difference, not a quotient.
• You must have a true quotient inside the logarithm. For example,
  $$
  \log_b\left(\frac{3x}{5}\right) = \log_b 3 + \log_b x - \log_b 5
  $$
  but
  $$
  \frac{\log_b 3x}{5}
  $$
  is a different expression; the 5 is not inside the log.

The power law

Statement

For any positive $M$, real number $k$, and base $b$,
$$
\log_b(M^k) = k \log_b M.
$$

So, an exponent inside the logarithm can be brought out front as a multiplier.

Why it matches exponent rules

If $b^x = M$, then $M^k = (b^x)^k = b^{kx}$. Taking $\log_b$,
$$
\log_b(M^k) = \log_b(b^{kx}) = kx = k \log_b M.
$$

Examples

• With base $10$:
  $$
  \log_{10}(100^3) = 3\log_{10} 100 = 3 \cdot 2 = 6.
  $$
• Symbolic:
  $$
  \log_5\big((x^2y)^3\big)
  = \log_5(x^6 y^3)
  = \log_5 x^6 + \log_5 y^3
  = 6\log_5 x + 3\log_5 y.
  $$

Working in reverse

The power law is often used backwards to combine logarithms:

• From $k \log_b M$ to $\log_b(M^k)$.

Example:
$$
2\log_3 x = \log_3(x^2), \quad
-\tfrac12 \log_7 y = \log_7(y^{-1/2}).
$$

Common mistakes

• The exponent must be on the argument of the log, not on the log itself:
• $\log_b(M^k)$ turns into $k\log_b M$.
• But $(\log_b M)^k$ is just the $k$‑th power of a number; there is no simplification rule for that.
• The exponent applies to the entire argument:
• $\log_b\big((3x)^2\big) = \log_b(9x^2)$, not $\log_b(3x^2)$.
• Then $\log_b(9x^2) = \log_b 9 + 2\log_b x$.

Special cases and identities

Some useful simple logarithm values and identities follow from the laws and from the definition of logarithms.

Logarithm of 1

For any base $b$,
$$
\log_b 1 = 0.
$$
Reason: $b^0 = 1$, so the exponent needed to get $1$ is $0$.

You can also get this from the quotient law:
$$
\log_b 1 = \log_b\left(\frac{b}{b}\right)
= \log_b b - \log_b b = 0.
$$

Logarithm of the base

For any base $b$,
$$
\log_b b = 1
$$
because $b^1 = b$.

Powers of the base

For any real $k$,
$$
\log_b(b^k) = k.
$$

This is just the definition of logarithm: $\log_b(b^k)$ is the exponent you put on $b$ to get $b^k$, which is $k$.

Using the power law in the opposite direction:
$$
k = \log_b(b^k) = \log_b\big((b)^k\big) = k \log_b b
$$
which is consistent with $\log_b b = 1$.

Logarithm of a root

A root is a fractional power, so combine the power law with that:
$$
\sqrt[n]{M} = M^{1/n}.
$$

Then
$$
\log_b\left(\sqrt[n]{M}\right) = \log_b(M^{1/n}) = \frac{1}{n}\log_b M.
$$

Examples:
$$
\log_2\sqrt{8} = \log_2(8^{1/2}) = \tfrac12 \log_2 8 = \tfrac12 \cdot 3 = \tfrac32.
$$

Expanding logarithmic expressions

To expand a logarithmic expression means to rewrite a single log of a complicated argument as a sum or difference of simpler logs using the product, quotient, and power laws.

General approach:

• Use the power law to move exponents out front.
• Use the product law to break up products inside a log into sums of logs.
• Use the quotient law to break up quotients inside a log into differences of logs.

Example 1

Expand:
$$
\log_2\left(\frac{8x^3}{y}\right).
$$

Step by step:
$$
\log_2\left(\frac{8x^3}{y}\right)
= \log_2(8x^3) - \log_2 y \quad\text{(quotient law)}
$$
$$
= \log_2 8 + \log_2 x^3 - \log_2 y \quad\text{(product law)}
$$
$$
= \log_2 8 + 3\log_2 x - \log_2 y \quad\text{(power law)}
$$
$$
= 3 + 3\log_2 x - \log_2 y \quad\text{(since } \log_2 8 = 3\text{)}.
$$

Example 2

Expand:
$$
\ln\left(\sqrt{x^2+1}\right).
$$

Recognize the root:
$$
\sqrt{x^2+1} = (x^2+1)^{1/2},
$$
so:
$$
\ln\left(\sqrt{x^2+1}\right)
= \ln\big((x^2+1)^{1/2}\big)
= \frac12 \ln(x^2+1).
$$

Note that $x^2+1$ is a sum; there is no log law to split $\ln(x^2+1)$ further.

Condensing logarithmic expressions

To condense (or combine) logarithmic expressions means to take a sum or difference of logs and rewrite it as a single log.

General approach:

• Use the power law in reverse to move coefficients into the log as exponents.
• Use the product law in reverse: sum of logs $\to$ log of a product.
• Use the quotient law in reverse: difference of logs $\to$ log of a quotient.

Important: You can only combine logs that have the same base and log of the same type (for example, all $\log_{10}$ or all $\ln$).

Example 1

Condense:
$$
\log_3 x + 2\log_3 y - \log_3 z.
$$

Step by step:

1. Use the power law in reverse:
   $$
   \log_3 x + 2\log_3 y - \log_3 z
   = \log_3 x + \log_3(y^2) - \log_3 z.
   $$
2. Combine the sum:
   $$
   \log_3 x + \log_3(y^2) = \log_3(xy^2).
   $$
3. Combine the difference:
   $$
   \log_3(xy^2) - \log_3 z = \log_3\left(\frac{xy^2}{z}\right).
   $$

So the condensed form is:
$$
\log_3\left(\frac{xy^2}{z}\right).
$$

Example 2

Condense:
$$
\frac12 \ln a - 3\ln b + \ln c.
$$

1. Use the power law in reverse:
   $$
   \frac12 \ln a = \ln(a^{1/2}), \quad
   3\ln b = \ln(b^3).
   $$
   So
   $$
   \frac12 \ln a - 3\ln b + \ln c
   = \ln(a^{1/2}) - \ln(b^3) + \ln c.
   $$
2. Combine $\ln(a^{1/2}) + \ln c$:
   $$
   \ln(a^{1/2}) + \ln c = \ln(a^{1/2}c).
   $$
3. Subtract $\ln(b^3)$:
   $$
   \ln(a^{1/2}c) - \ln(b^3)
   = \ln\left(\frac{a^{1/2}c}{b^3}\right).
   $$

Domain issues when applying logarithmic laws

The logarithmic laws assume that all arguments of the logarithms are positive. When working with variables, this matters.

Examples:

• From $\log_b(x^2)$ to $2 \log_b x$:
• $\log_b(x^2)$ is defined for all $x \ne 0$ (since $x^2>0$ when $x\ne 0$).
• $2 \log_b x$ is defined only for $x>0$.

As functions, these are not equal on the whole real line, though in many algebraic manipulations (especially when solving equations with the understanding that $x>0$) we still use the law. Be aware of the domain restriction.

• From $\log_b(-3x)$:
• This expression is only defined when $-3x > 0$, i.e., when $x < 0$.
• If you rewrite $\log_b(-3x) = \log_b 3 + \log_b(-x)$, then $\log_b(-x)$ requires $-x>0$, which again means $x<0$. The domain is consistent, but you must keep track of sign conditions.

In typical Algebra II problems, you often assume the variables take values that keep every logarithm defined (positive arguments), but when solving equations, you must still check that your solutions make every log argument positive.

Summary of logarithmic laws

For $b>0$, $b\ne 1$, and positive $M, N$:

• Product law
  $$
  \log_b(MN) = \log_b M + \log_b N
  $$
• Quotient law
  $$
  \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N
  $$
• Power law
  $$
  \log_b(M^k) = k \log_b M
  $$
• Special values
  $$
  \log_b 1 = 0, \quad \log_b b = 1, \quad \log_b(b^k) = k.
  $$

These laws let you expand a single complicated logarithm into simpler pieces, or condense several logarithms into a single one. They will be used repeatedly when simplifying expressions and when solving logarithmic and exponential equations.