1. **State the problem:** Solve the equation $2^x + 3x = 17$ for $x$. 2. **Understand the equation:** This is a transcendental equation involving both an exponential term $2^x$ and a linear term $3x$. Such equations often cannot be solved exactly using elementary algebraic methods. 3. **Approach:** We can try to find solutions by testing integer values or use numerical methods like the Newton-Raphson method. 4. **Test integer values:** - For $x=2$: $2^2 + 3(2) = 4 + 6 = 10$ (less than 17) - For $x=3$: $2^3 + 3(3) = 8 + 9 = 17$ (equals 17) 5. **Check if $x=3$ is a solution:** Substitute $x=3$ back into the equation: $$2^3 + 3 \times 3 = 8 + 9 = 17$$ This satisfies the equation. 6. **Check for other solutions:** - For $x=4$: $2^4 + 3(4) = 16 + 12 = 28$ (greater than 17) - For $x=1$: $2^1 + 3(1) = 2 + 3 = 5$ (less than 17) Since the function $f(x) = 2^x + 3x$ is continuous and increasing for $x > 0$, and we found $f(2) < 17$ and $f(3) = 17$, the only solution in this range is $x=3$. 7. **Final answer:** $$x = 3$$