Exponential equations and Bernoulli’s inequality: signs relative to a chord

Content updated: October 22, 2026

An exponential graph lies below the chord joining its points at 0 and 1, and above the extensions of that chord outside the interval. This sign relation determines the exact solutions of many exponential equations with a linear right side.

Objectives: apply Bernoulli's inequality on the correct intervals; prove that there are no roots beyond two known points; combine expressions having the same sign.

Prerequisites: exponential functions, second derivatives, and strict convexity.

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

1. Statement and boundary points

For $a>0$, $a\ne1$, set $L(x)=1+(a-1)x$. This line passes through $(0;1)$ and $(1;a)$. The real-exponent form of Bernoulli's inequality is

Nonstrict inequalities may be combined on the corresponding closed intervals, but the strict less-than sign fails at the endpoints 0,1. For $a=1$, both the exponential and the line equal 1, so equality holds for every $x$; the hypothesis $a\ne1$ is needed to assert just two equality points.

2. Why the inequality holds

The function $f(x)=a^x$ has $f^{\prime\prime}(x)=a^x(\ln a)^2>0$, so it is strictly convex. Hence for $0<x<1$, its value is less than the combination $(1-x)f(0)+xf(1)=L(x)$. This is the chord property of a strictly convex function.

For $x<0$, write $0=\lambda x+(1-\lambda)1$, where $\lambda=1/(1-x)\in(0;1)$. Strict convexity implies $1<\lambda a^x+(1-\lambda)a$; solving for $a^x$ gives $a^x>L(x)$. For $x>1$, write $1=(1-1/x)0+(1/x)x$ and argue in the same way. Equality at 0,1 follows from the definition of $L$.

These arguments also apply for $0<a<1$, since the second derivative remains positive. For equations, group each difference $a^x-L(x)$ as one expression. If several such expressions have the same sign on each region, their sum cannot be 0 away from their common equality points.

3. Worked examples

Example 1. A line through two points of 5ˣ

Solve $5^x-4x-1=0$. For $a=5$, we have $L(x)=4x+1$. The difference $5^x-L(x)$ is positive for $x<0$ or $x>1$, negative for $0<x<1$, and 0 at 0,1. Therefore the equation has exactly two solutions, $x=0,x=1$.

Example 2. Two differences with matching signs

Solve $3^x+4^x=5x+2$. Rearrange as $(3^x-2x-1)+(4^x-3x-1)=0$. Each term is a Bernoulli difference. Both are positive outside $[0;1]$ and negative on $(0;1)$. Their sum is 0 only at the endpoints, where both vanish. Thus the solutions are 0 and 1.

This argument does not assert that the terms are always nonnegative: they are negative in the middle interval. What matters is that they have the same sign at each $x$ and vanish together only at the two specified points.

Example 3. A base between 0 and 1

Solve $(1/2)^x=1-x/2$. For $a=1/2$, the right side is exactly $L(x)$. The sign relative to the chord follows the same Bernoulli rule, so the solutions remain 0 and 1. Where the right side is nonpositive, direct exclusion is also possible because the left side is always positive.

4. Practice exercises

1. Solve $2^x=x+1$.
2. Solve $2^x+5^x=5x+2$.
3. For $a=1$, how many solutions does $a^x=1+(a-1)x$ have?

5. Common mistakes

Do not use a strict less-than sign on the closed interval [0;1]. Retain the base-1 exception. When combining inequalities, check that the differences really have the same sign at the same x; opposite-sign expressions can cancel without either being 0.

6. Related lessons

• Exponential functions (English lesson)
• Derivatives of exponential and logarithmic functions (English lesson)
• Exponential equations using function properties (English lesson)
• Exponential equations using a common base (English lesson)

7. Material accompanying the original lesson

• Scribd exercise material from the original lesson 1
• Enrollment link from the original lesson