Turning an Equation into a Root Problem
An equation involving an exponential, logarithmic, or trigonometric function may be difficult to solve exactly. Even when an exact solution is unavailable, it can still be possible to prove that at least one solution exists.
The key is to rewrite the equation as a zero of a function. For example, the equation \(e^x=2\) is equivalent to \(e^x-2=0\). If the resulting function is continuous and takes values of opposite signs at two endpoints, the Intermediate Value Theorem guarantees a zero between them.
The Difference-Function Method
Suppose the equation to be considered is \(f(x)=g(x)\). Define the difference function
The original equation holds at \(x=c\) exactly when \(H(c)=0\), because
This converts an equation into a root-finding question. The function \(H\) must be continuous on the entire interval being used, not just at its endpoints. Once that condition and an endpoint sign change are verified, the Intermediate Value Theorem applies.
The strict inequality \(H(a)H(b)<0\) means that one endpoint value is positive and the other is negative. If an endpoint value is already zero, then that endpoint itself is a solution, so no sign-change argument is needed.
Why the Method Works
The central reason is that the difference of continuous functions is continuous. Here is the justification that connects the equation to the theorem.
For the equation \(f(x)=g(x)\), set \(H(x)=f(x)-g(x)\). The solutions of the equation are precisely the zeros of \(H\).
If \(f\) and \(g\) are continuous throughout \([a,b]\), then their difference \(H=f-g\) is continuous throughout \([a,b]\).
If \(H(a)H(b)<0\), then \(0\) lies strictly between \(H(a)\) and \(H(b)\).
Continuity ensures that \(H\) takes every intermediate value between \(H(a)\) and \(H(b)\). In particular, \(H(c)=0\) for some \(c\in(a,b)\), so \(f(c)=g(c)\).
This explains why the signs matter: the endpoint values straddle zero, and a continuous function cannot pass from a negative value to a positive value without taking the value zero along the way. The argument proves existence; it does not generally give the exact location of \(c\).
Worked Examples
Worked Example: An Exponential Equation
Prove that \(e^x=2\) has a solution in \((0,1)\), without solving the equation.
Define the difference function
The exponential function is continuous for every real \(x\), so \(H\), a difference of continuous functions, is continuous on \([0,1]\). At the left endpoint,
At the right endpoint,
since \(e>2\). Thus
By the Intermediate Value Theorem, some \(c\in(0,1)\) satisfies \(H(c)=0\). Substituting the definition of \(H\), \(e^c-2=0\), hence \(e^c=2\). This verifies that the zero of the difference function solves the original equation.
Worked Example: A Logarithmic Equation
Prove that \(\ln x=2-x\) has a solution in \((1,2)\).
Move the right-hand side to the left and define
The logarithm \(\ln x\) is defined and continuous for \(x>0\). In particular, \(H\) is continuous throughout the closed interval \([1,2]\). Evaluate both endpoints:
Therefore \(H(1)H(2)=(-1)(\ln 2)<0\). The Intermediate Value Theorem gives a \(c\in(1,2)\) with \(H(c)=0\). That equation says \(\ln c+c-2=0\), or \(\ln c=2-c\), as required.
Worked Example: A Trigonometric Equation
Prove that \(\sin x=\frac12\) has a solution in \((0,\frac{\pi}{2})\).
Define
The sine function is continuous for every real \(x\), so \(H\) is continuous on \([0,\frac{\pi}{2}]\). At the endpoints,
Their product is \((-\frac12)(\frac12)=-\frac14<0\). By the Intermediate Value Theorem, there is some \(c\in(0,\frac{\pi}{2})\) such that \(H(c)=0\). Substituting back gives \(\sin c-\frac12=0\), so \(\sin c=\frac12\).
Choosing an Interval and Checking the Domain
The interval is part of the proof, not a detail to add afterward. It must meet two requirements: the difference function must be defined and continuous at every point of the closed interval, and its endpoint values must have opposite signs (unless an endpoint is already a solution).
For exponential and trigonometric functions, the functions themselves are defined and continuous for all real inputs. The other terms in the equation can still restrict the domain. Logarithmic expressions require particular care: \(\ln x\) is defined only when \(x>0\). If a denominator occurs, its value must also remain nonzero throughout the chosen interval.
A useful way to organize a proof is to check these points in order:
- Write the equation as \(H(x)=0\), with all terms included in \(H\).
- Choose a closed interval \([a,b]\) on which every expression in \(H\) is defined.
- Verify that \(H\) is continuous at every point of that interval.
- Calculate \(H(a)\) and \(H(b)\), keeping their signs exact where possible.
- Apply the Intermediate Value Theorem and translate \(H(c)=0\) back into the original equation.
Endpoint calculations should be shown explicitly. Saying that the signs are opposite without evaluating the function, or without justifying the sign of an expression such as \(\ln 2\), leaves a gap in the argument.
Common Pitfall: Using an Interval Outside the Domain
Consider again \(H(x)=\ln x+x-2\). A tempting but invalid argument might use \([0,2]\), calculate an endpoint value at \(2\), and claim continuity there and across the interval. But \(\ln 0\) is undefined, so \(H(0)\) does not exist. The function is not defined, and hence is not continuous, throughout \([0,2]\). The Intermediate Value Theorem cannot be applied on that interval.
The correction is to choose an interval contained in the domain, such as \([1,2]\). On this interval \(x>0\), so the logarithm is defined and continuous throughout. The endpoint values are \(H(1)=-1<0\) and \(H(2)=\ln 2>0\), which gives a valid sign change and proves existence.
What the Conclusion Does—and Does Not—Say
The conclusion is that at least one solution lies in the open interval. It does not usually identify the solution exactly, and one sign change does not by itself establish that there is only one solution. For this method, the essential conclusion is existence.
The argument is useful when an equation is hard to solve explicitly but its terms are continuous and their difference is easy to evaluate at selected endpoints. Exponentials, logarithms, and trigonometric functions often appear in models where exact formulas are unavailable. A carefully chosen interval can still give a rigorous guarantee that the model reaches a specified value.
Check Your Understanding
For each question, identify the difference function, its domain condition, and the endpoint sign information needed for an IVT argument.
- To study \(e^x=3\), what difference function could you define?
- Why is \([0,1]\) not a valid interval for applying IVT to \(H(x)=\ln x-1\)?
- If \(H\) is continuous on \([a,b]\), \(H(a)<0\), and \(H(b)>0\), what does IVT guarantee?
- For the equation \(\cos x=x\), write a suitable difference function. What must be checked before using IVT on a proposed interval?
- Does proving a sign change for a continuous difference function establish that the equation has exactly one solution? Explain what the sign-change argument establishes.
