Logarithm Laws

Fix a valid base and positive arguments

For real logarithms, the base satisfies $b>0$ and $b\ne1$. The argument must be strictly positive. The definition is

A base between zero and one is valid, although its logarithm is decreasing. Base one is not valid because $1^u$ cannot distinguish exponents. In this page $\ln$ means base $e$, and $\log_{10}$ is written with its base whenever a decimal logarithm is intended.

The inverse relations are

These do not give a real logarithm of zero or a negative number. Complex logarithms have branch issues and are outside these real-variable rules.

Derive the product and quotient laws

Let $u=\log_b M$ and $v=\log_b N$, with $M,N>0$. Then $M=b^u$ and $N=b^v$. The exponent laws give

Applying the inverse function yields

For example,

No decimal approximations are necessary. Both logarithms on the left are defined, and the quotient law preserves their value.

The positivity requirement is not redundant. The number $(-2)(-3)=6$ has a real logarithm, but neither $\ln(-2)$ nor $\ln(-3)$ does. Thus $\ln(MN)$ can have a larger real domain than $\ln M+\ln N$. An identity derived under $M,N>0$ does not justify adding other cases silently.

Derive the power law and use absolute values when needed

For $M>0$ and real $r$, write $M=b^u$. Then $M^r=b^{ur}$ and

For instance, $\log_3(9^2)=2\log_3 9=4$. The rule follows from exponent structure, not from moving a printed exponent by convention.

For a variable that might be negative, extra care is required. The expression $\ln(x^2)$ is defined whenever $x\ne0$. Since $x^2=|x|^2$,

Writing $2\ln x$ would preserve the value only on $x>0$ and would lose every negative input from the original domain. Likewise, for a positive integer $m$,

The OpenStax logarithmic-properties chapter gives the standard laws. The domain comparisons here explain how to use them when the original factors do not all have a positive sign.

Expand a compound expression with its full domain

Consider

For a real square root and a positive logarithm argument, the original domain requires $x\ne0$, $y>0$ and $z>0$. If $y=0$ or $x=0$, the argument is zero; if $z<0$, it is negative; if $z=0$, the quotient is undefined.

On that domain,

Check a permitted negative value of $x$: with $x=-2$, $y=9$, $z=1$, the original argument is 12 and the expansion is $2\ln2+\ln3=\ln12$. The absolute value is necessary for the expanded expression to remain defined.

This example also shows why domain analysis should come before the formal expansion. A denominator’s odd power and a square root constrain signs differently.

Condense without enlarging the original domain

For

the original domain is $x>1$. The power and quotient laws give

The retained condition makes the equivalence explicit. In other examples, the condensed argument may be positive on additional intervals, even though the separate logarithms are undefined there.

For example, $\ln(x-1)+\ln(x+1)$ has domain $x>1$, whereas $\ln(x^2-1)$ also exists for $x<-1$. Condensing an equation built from the two original logarithms does not authorise those extra negative inputs. Solving Logarithmic Equations applies this distinction to extraneous roots.

Derive change of base

Let $u=\log_b x$, so $b^u=x$. For another valid base $a$,

Since $b\ne1$, the denominator is nonzero, giving

This allows a calculator with a natural-logarithm button to evaluate any valid positive base. Keep the argument above the base logarithm: $\ln b/\ln x$ is generally a different quantity.

When $x>0$ and $x\ne1$, reversing the base and argument gives

For example, $\log_2 8=3$ and $\log_8 2=1/3$. At $x=1$, the second logarithm would have the invalid base one, so the product relation does not apply.

Identify operations for which no corresponding law exists

There is no general sum law $\ln(M+N)=\ln M+\ln N$. The right side equals $\ln(MN)$ when $M,N>0$, not the logarithm of their sum. For $M=1$ and $N=2$, these arguments are 3 and 2, so the proposed equality fails.

Likewise, $\ln(M-N)$ cannot generally be split into a difference of logarithms. That difference represents $\ln(M/N)$. A useful strategy is to factor a sum or difference first, then apply a law only to the resulting valid product or quotient.

The square of a logarithm, $(\ln x)^2$, is also different from $\ln(x^2)$. At $x=e$, the first equals 1 and the second equals 2. Parentheses determine which operation is being performed.

Related topics

Use Solving Exponential Equations to bring an unknown exponent into an equation. Solving Logarithmic Equations develops domain-sensitive solutions and inequalities.