Solving Logarithmic Equations

Establish the domain before choosing a method

For a fixed real base $b>0$, $b\ne1$, an expression $\log_b U(x)$ requires $U(x)>0$. When several logarithms appear, all their conditions must hold at once.

For example, $\log_3(x-1)+\log_3(x+1)$ requires both $x>1$ and $x>-1$, so the domain is $x>1$. The product $(x-1)(x+1)$ is also positive for $x<-1$, but that does not extend the domain of the original sum.

This distinction is why writing domain restrictions only after condensing can introduce extraneous solutions. Logarithm Laws explains the underlying identities and their conditions.

Convert a single logarithm to an exponential equation

Solve

The domain is $x>-1$. The logarithm definition gives $x+1=5^2=25$, so $x=24$. This is permitted, and the original logarithm is $\log_5 25=2$.

Now consider

The original condition is $x^2-4>0$, or $x<-2$ or $x>2$. Exponentiating gives $x^2-4=8$, so

Both candidates satisfy the domain. A logarithmic equation can have more than one real solution when its argument is not one-to-one. The logarithm itself is one-to-one as a function of a positive argument; that does not make every composite argument one-to-one in $x$.

Combine logarithms and filter the algebraic roots

Solve

On its domain $x>1$, use the product law:

The algebraic candidates are $x=\pm\sqrt{10}$. Only $\sqrt{10}$ exceeds 1. Therefore the complete solution set is $\{\sqrt{10}\}$.

The negative candidate would make both original arguments negative. It is not a solution even though their product is positive. It belongs to the expanded domain of the condensed expression, not the original equation.

For another example,

requires $x>2$. The transformed equation is $x(x-2)=8$, or $(x-4)(x+2)=0$. The candidates are 4 and −2. Keep 4, for which the original logarithms are 2 and 1, and reject −2.

Equate arguments only with matching valid bases

For a fixed valid base, $\log_b U=\log_b V$ is equivalent to $U=V$ provided both arguments are positive.

Solve

The domain is $2<x<6$. Equality of arguments gives $x-2=6-x$, hence $x=4$, which is permitted. Both sides then equal $\ln2$.

With different bases, equal logarithms do not imply equal arguments. For example, $\log_2 8=\log_3 27=3$, but $8\ne27$. Convert bases or use the definitions instead of dropping unlike logarithms.

The OpenStax logarithmic-equation chapter presents the definition and one-to-one methods used here.

Treat a repeated logarithm as a new variable

Solve

The original domain is $x>0$. Let $u=\ln x$. Unlike an exponential substitution such as $u=2^x$, this $u$ can be any real number. Factor:

Thus $\ln x=1$ or $\ln x=2$, giving $x=e$ or $x=e^2$. Both are positive. Returning to the original variable is necessary; the numbers 1 and 2 are logarithm values, not the final values of $x$.

Do not confuse $(\ln x)^2$ with $\ln(x^2)$. They are different functions and lead to different equations.

Include base restrictions when the base is variable

Solve

A real variable base requires $x>0$ and $x\ne1$. The logarithm definition gives $x^2=16$, producing algebraic candidates $x=\pm4$. Only 4 is a permitted base, and $\log_4 16=2$.

The restriction is not optional because 16 is positive. Positivity of the argument and validity of the base are separate conditions. A base-one candidate must also be rejected, even if a transformed exponential equation happens to include it.

For instance, $\log_x x=1$ holds for every $x>0$, $x\ne1$, not at every positive input without exception.

Reverse order for a decreasing logarithm

Solve

First require $x>1$. Since the base $1/2$ lies between zero and one, its logarithm is decreasing. Also, $-2=\log_{1/2}4$. Therefore

which gives the solution interval

The lower endpoint comes from the domain, not from solving an equality. The upper endpoint is excluded because the inequality is strict. For a logarithm base greater than one, argument inequalities retain their direction.

Check transformations and completeness

Exponentiating two equal real numbers with the same valid base is reversible, so it does not inherently create extra real solutions. Problems arise when a transformed expression is used beyond the original domain, or when a many-to-one operation such as squaring is applied without checking candidates.

A complete solution states the original domain, gives transformations valid there, solves the remaining algebra, and filters all candidates. Substituting the survivors checks arithmetic. The chain of equivalences explains why no permitted solution was lost.

Graphing helps locate candidates or illustrate monotonicity, but a displayed curve cannot decide an exact endpoint or prove that no other branch exists. Use algebra and domain reasoning for those decisions.

Related topics

Use Exponential Growth and Decay for model fitting and thresholds. Composite and Inverse Functions: Advanced Problems extends domain analysis to inverse formulas and multi-stage functions.