Find the sum of all the solutions to $2^{|x|} + 3|x| = 18.$
If $x$ is a solution, then $-x$ is a also a solution.  Thus, we can pair all the solutions, and their sum is $\boxed{0}.$

Let $f(x) = 2^{|x|} + 3|x|.$  Since $f(0) = 0$ and $f(4) = 28,$ the equation $f(x) = 18$ has at least one solution in the interval $0 \le x \le 4.$  This ensures that the sum that the problem asks for is not an "empty" sum.
0
Level 2
Intermediate Algebra
Problem: Find the sum of all the solutions to $2^{|x|} + 3|x| = 18.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation \(2^{|x|} + 3|x| = 18\), we first isolate \(2^{|x|}\) by rewriting the equation as \(2^{|x|} = 18 - 3|x|\). This requires \(18 - 3|x| \geq 0\), leading to the condition \(|x| \leq 6\).

Next, we analyze the function \(f(|x|) = 2^{|x|} + 3|x|\). We need to find the values of \(|x|\) that satisfy \(f(|x|) = 18\). The function \(f(|x|)\) is continuous and strictly increasing for \(|x| \geq 0\) because both components \(2^{|x|}\) and \(3|x|\) are increasing functions.

We evaluate \(f(0)\) and \(f(6)\):
- \(f(0) = 2^0 + 3 \cdot 0 = 1\)
- \(f(6) = 2^6 + 3 \cdot 6 = 64 + 18 = 82\)

Since \(f(0) < 18 < f(6)\), by the Intermediate Value Theorem, there exists at least one solution in the interval \((0, 6)\).

Next, we can check specific values to narrow down the solutions:
- For \(|x| = 2\): \(f(2) = 2^2 + 3 \cdot 2 = 4 + 6 = 10\)
- For \(|x| = 3\): \(f(3) = 2^3 + 3 \cdot 3 = 8 + 9 = 17\)
- For \(|x| = 3.5\): \(f(3.5) = 2^{3.5} + 3 \cdot 3.5 \approx 11.31 + 10.5 = 21.81\)

The solution lies between \(|x| = 3\) and \(|x| = 3.5\). We can use numerical methods or further evaluations to find the exact solutions. 

Assuming we find two solutions, \(|x| = a\) and \(|x| = b\), the sum of all solutions is \(a + (-a) + b + (-b) = 0\). Thus, the final answer is \(0\).<|END_OF_LATENT|> The final answer is $0$.