Exponentials Logarithms 571F74

1. **Problem 1:** Find the sum of the real solutions to the equation $$2^x - (\sqrt{2})^{x+6} + 12 = 0.$$ 2. **Rewrite the equation:** Note that $\sqrt{2} = 2^{\frac{1}{2}}$, so $$(\sqrt{2})^{x+6} = \left(2^{\frac{1}{2}}\right)^{x+6} = 2^{\frac{x+6}{2}} = 2^{\frac{x}{2} + 3} = 2^3 \cdot 2^{\frac{x}{2}} = 8 \cdot 2^{\frac{x}{2}}.$$ 3. **Substitute back:** The equation becomes $$2^x - 8 \cdot 2^{\frac{x}{2}} + 12 = 0.$$ 4. **Use substitution:** Let $$y = 2^{\frac{x}{2}}.$$ Then $$2^x = \left(2^{\frac{x}{2}}\right)^2 = y^2.$$ 5. **Rewrite the equation in terms of $y$:** $$y^2 - 8y + 12 = 0.$$ 6. **Solve the quadratic:** Use the quadratic formula: $$y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}.$$ 7. **Find roots for $y$:** $$y_1 = \frac{8 + 4}{2} = 6, \quad y_2 = \frac{8 - 4}{2} = 2.$$ 8. **Back-substitute for $x$:** Recall $$y = 2^{\frac{x}{2}}.$$ So $$2^{\frac{x}{2}} = 6 \implies \frac{x}{2} = \log_2 6 \implies x = 2 \log_2 6,$$ and $$2^{\frac{x}{2}} = 2 \implies \frac{x}{2} = \log_2 2 = 1 \implies x = 2.$$ 9. **Sum of real solutions:** $$x_1 + x_2 = 2 \log_2 6 + 2 = 2(1 + \log_2 3) + 2 = 2 + 2 + 2 \log_2 3 = 4 + 2 \log_2 3.$$ 10. **Answer for Problem 1:** The sum of the real solutions is $$4 + 2 \log_2 3,$$ which corresponds to option F. --- 11. **Problem 2:** Given $$\frac{27^{2(x-2)}}{9^{2(x-3)}} = 81^{\frac{3}{2}},$$ find the value of $x$. 12. **Rewrite bases as powers of 3:** $$27 = 3^3, \quad 9 = 3^2, \quad 81 = 3^4.$$ 13. **Rewrite the equation:** $$\frac{(3^3)^{2(x-2)}}{(3^2)^{2(x-3)}} = (3^4)^{\frac{3}{2}}.$$ 14. **Simplify exponents:** $$\frac{3^{6(x-2)}}{3^{4(x-3)}} = 3^{6}.$$ 15. **Use quotient rule:** $$3^{6(x-2) - 4(x-3)} = 3^{6}.$$ 16. **Simplify exponent:** $$6(x-2) - 4(x-3) = 6x - 12 - 4x + 12 = 2x.$$ 17. **Equate exponents:** $$2x = 6 \implies x = 3.$$ 18. **Answer for Problem 2:** The value of $x$ is $$3.$$