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Logarithmic equations using monotonicity: counting roots on the correct domain

2026-10-22 huynhvietkhanh Comments: 0

Use monotonicity and Rolle’s theorem on each domain interval, distinguishing uniqueness bounds from existence.

High schoolPowers & Logarithms

Content updated: October 22, 2026

Monotonicity can prove that a logarithmic solution is unique without reducing the equation to algebra. A root bound applies only on an interval contained in the domain; a domain split into several intervals must be treated piece by piece.

Objectives: form an auxiliary function with correct signs and constants; prove completeness of the solutions; recognize the limits of monotonicity arguments on a disconnected domain.

Prerequisites: logarithmic derivatives, monotonicity, limits, and Rolle's theorem.

On this page: Idea · Method · Examples · Practice · Further learning

1. Derivatives and uniqueness

For $a>0$, $a\ne1$, and $x>0$, the derivative of $\log_a x$ is $1/(x\ln a)$. Thus the logarithm increases for $a>1$ and decreases for $0<a<1$. Determine the domains of all arguments and bases before examining derivative signs.

If $F$ is differentiable and $F^{\prime}>0$ or $F^{\prime}<0$ on an interval $I$, the equation $F(x)=0$ has at most one solution in $I$. To conclude uniqueness, verify an $x_0$ in $I$ satisfying $F(x_0)=0$. A nondecreasing function may vanish on an entire interval, so nonstrict derivative signs require care.

2. Rolle's theorem and domains with several intervals

If $F$ has derivatives through order $k\ge1$ on an interval $I$ and $F^{(k)}$ never vanishes on $I$, then $F$ has at most $k$ distinct roots in $I$. Otherwise, $k+1$ roots and repeated applications of Rolle's theorem would produce a zero of the derivative of order $k$, contradicting the hypothesis.

When rewriting an equation as $F=0$, retain the correct signs of every term and constant. Then prove monotonicity directly or compare an increasing side with a decreasing side. If no root has been found, “at most one” does not guarantee existence; verify a root or additionally use continuity and a sign change.

On a domain such as $(-\infty;0)\cup(0;+\infty)$, Rolle's theorem cannot be applied across 0 when the function is undefined there. Each interval may have its own root. A derivative formula does not remove the need to examine the domain's structure.

3. Worked examples

Example 1. An equation with base one-half

Solve $\log_{1/2}x=5x-3/2$. Require $x>0$. Substitution of $x=1/2$ makes both sides 1. For uniqueness, set $F(x)=\log_{1/2}x-5x+3/2$ on $(0;+\infty)$. Then

$$F^{\prime}(x)=\frac1{x\ln(1/2)}-5=-\frac1{x\ln2}-5<0.$$

Thus $F$ is strictly decreasing, so $x=1/2$ is the unique solution. Alternatively, without derivatives, the base-1/2 logarithm decreases while the line $5x-3/2$ increases, so there is at most one intersection; the verified solution proves that it exists.

Example 2. An increasing function

Solve $\ln x+x=1$, with $x>0$. Set $G(x)=\ln x+x-1$; then $G^{\prime}(x)=1/x+1>0$. Since $G(1)=0$, the unique solution is $x=1$. No general algebraic solution formula for all equations involving ln and x is needed.

Example 3. Two intervals yielding two roots

Solve $\log_2(x^2)=0$. The domain is $x\ne0$. The equation is equivalent to $x^2=1$, giving $x=-1$ and 1. Its derivative is $2/(x\ln2)$: negative on the negative interval and positive on the positive interval, so each interval has at most one root. The two roots do not contradict monotonicity on each interval; the intervals must not be joined across the excluded point 0.

4. Practice exercises

  1. Solve $\log_3 x+x=1$.
  2. Prove that $\ln x=x-1$ has the unique solution 1.
  3. Solve $\log_{1/2}(x^2)=0$ and specify the intervals to examine if using derivatives.
Show hints and answers
  1. The domain is $x>0$. The function $F=\log_3 x+x-1$ has $F^{\prime}=1/(x\ln3)+1>0$ and $F(1)=0$, so the solution is 1.
  2. Set $H=\ln x-x+1$ on the positive domain. Its derivative $1/x-1$ is positive for $0<x<1$ and negative for $x>1$, so $H$ has its unique maximum at $x=1$, with $H(1)=0$. Thus only $x=1$ satisfies $H(x)=0$.
  3. The domain is $x\ne0$, and $x^2=1$ gives $-1,1$. Examine derivatives separately on $(-\infty;0)$ and $(0;+\infty)$; do not apply the argument across 0.

5. Common mistakes

Do not use an auxiliary function differing from the equation through an incorrect constant. A bound on the number of roots is not an existence proof. Do not infer uniqueness on the whole domain from monotonicity on separate intervals. Check solutions against the original arguments and bases.

6. Related lessons

7. Videos and reference material

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