1. **State the problem:** Solve the equation $x^x = 49$ for $x$. 2. **Understand the equation:** The equation $x^x = 49$ means we want to find a number $x$ such that when $x$ is raised to the power of itself, the result is 49. 3. **Rewrite 49 as a power:** Note that $49 = 7^2$. 4. **Try to express $x$ in terms of 7:** Suppose $x = 7^a$ for some $a$. Then: $$x^x = (7^a)^{7^a} = 7^{a \cdot 7^a}$$ We want this to equal $7^2$, so: $$7^{a \cdot 7^a} = 7^2$$ 5. **Equate exponents:** Since the bases are the same and nonzero, the exponents must be equal: $$a \cdot 7^a = 2$$ 6. **Solve for $a$:** This is a transcendental equation and does not have a simple algebraic solution. We can try to find $a$ approximately. 7. **Check integer values:** - For $a=0$, $0 \cdot 7^0 = 0$ (too small) - For $a=1$, $1 \cdot 7^1 = 7$ (too large) 8. **Try a value between 0 and 1:** - For $a=0.3$, $0.3 \cdot 7^{0.3} \approx 0.3 \cdot 1.903 = 0.571$ (too small) - For $a=0.5$, $0.5 \cdot 7^{0.5} = 0.5 \cdot \sqrt{7} \approx 0.5 \cdot 2.6458 = 1.3229$ (still too small) - For $a=0.7$, $0.7 \cdot 7^{0.7} \approx 0.7 \cdot 3.727 = 2.609$ (too large) 9. **Interpolate:** The value of $a$ is between 0.5 and 0.7. By linear approximation, $a \approx 0.6$. 10. **Find $x$:** Recall $x = 7^a$, so: $$x \approx 7^{0.6} = e^{0.6 \ln 7} \approx e^{0.6 \times 1.9459} = e^{1.1675} \approx 3.213$$ 11. **Check:** $$3.213^{3.213} \approx e^{3.213 \ln 3.213} \approx e^{3.213 \times 1.1675} = e^{3.75} \approx 42.5$$ This is close but less than 49, so $a$ should be slightly larger. 12. **Refine estimate:** Using a calculator or numerical methods, the solution is approximately: $$x \approx 3.3$$ **Final answer:** $$x \approx 3.3$$