Solve the equation 2x = 2^x.

Step-by-step solution

1. 1

   Test small values first. The equation mixes a linear and an exponential term, so no algebraic rearrangement will separate $x$. Substituting integers:

   $x=1:\ 2(1)=2\ \text{and}\ 2^{1}=2\ \checkmark$

   $x=2:\ 2(2)=4\ \text{and}\ 2^{2}=4\ \checkmark$

   Both are exact solutions ✓. The question is then whether any others exist.

2. 2

   Reframe as a single function. Let

   $f(x)=2^{x}-2x$

   Solutions of the original equation are exactly the zeros of $f$. We already know $f(1)=0$ and $f(2)=0$.

3. 3

   Use convexity to cap the number of roots. The function $2^{x}$ is convex — its second derivative $(\ln2)^{2}2^{x}$ is positive everywhere — while $2x$ is a straight line. A line can cross a strictly convex curve at most twice. Since two crossings are already in hand, there can be no more.

4. 4

   Confirm the shape by sampling. $f(0)=1-0=1>0$, $f(1.5)=2^{1.5}-3\approx2.828-3=-0.172<0$, and $f(3)=8-6=2>0$ ✓. The pattern positive, negative, positive shows $f$ dips below zero exactly between $x=1$ and $x=2$ and stays positive outside — precisely two sign changes, hence exactly two roots.

5. 5

   State the complete solution.

   $x=1\qquad\text{or}\qquad x=2$

   Note how special this is: $x=1$ solves $b^{x}=bx$ for every base $b$, since both sides equal $b$. What singles out base $2$ is the second crossing: $b^{2}=2b$ forces $b=2$, so no other base has an integer partner. For $b=3$, say, the second root is the irrational $x\approx 0.826018$.