Solving equations with an added straight line

Some equations, such as \(2\ln x+x=4\), cannot be solved by algebra alone. If one side of a rearranged equation is a curve you have sketched, the other side can be drawn on the same axes as a straight line. Each point where the line meets the curve gives one solution.

Why: a point lies on both graphs exactly when its coordinates satisfy both equations, so its \(x\)-coordinate makes the two sides equal.

Method

1. Rearrange the equation so that one side is exactly the function already sketched, for example \(\ln x=\ldots\) or \(\mathrm e^{2x}=\ldots\).
2. The other side should be a straight line \(y=mx+c\). State its equation.
3. Draw the line using two points, such as its intercepts.
4. Count the intersections for the number of solutions; read an \(x\)-coordinate if an estimate is asked for.
5. Check the domain: a logarithm needs a positive argument, and an equation with \(x^2\) may have solutions that the sketched branch cannot show.

Example

Show how the graph of \(y=\ln x\) can be used to find the number of solutions of \(2\ln x+x=4\).

\(\begin{aligned} 2\ln x&=4-x \\ \ln x&=2-\tfrac12x \end{aligned}\)

Draw the line \(y=2-\tfrac12x\), through \((0,\,2)\) and \((4,\,0)\), on the sketch of \(y=\ln x\).

\(y=\ln x\) increases and the line decreases, so they meet once: the equation has one solution, \(x\approx2.3\).

Counting by changes of side

Each change of side between the curve and the line marks a crossing: check which is higher at a few chosen values of \(x\).

A curve that bends one way throughout meets a straight line at most twice: \(y=a^x\) always bends upwards; \(y=\log_a x\) bends downwards when \(a>1\) (such as \(\ln x\) and \(\lg x\)) and upwards when \(0<a<1\); a negative coefficient turns the bend over. So two changes of side on such a curve give exactly two solutions.