1. **State the problem:** Solve the exponential equation $$5^{2x+4} + 1 = 26 \cdot 5^{x+1}$$. 2. **Recall the properties of exponents:** - $a^{m+n} = a^m \cdot a^n$ - $a^{2x} = (a^x)^2$ 3. **Rewrite terms to have the same base and simplify:** $$5^{2x+4} = 5^{2x} \cdot 5^4 = (5^x)^2 \cdot 625$$ $$26 \cdot 5^{x+1} = 26 \cdot 5^x \cdot 5 = 130 \cdot 5^x$$ 4. **Substitute $y = 5^x$ (since $5^x > 0$):** $$625 y^2 + 1 = 130 y$$ 5. **Rearrange to form a quadratic equation:** $$625 y^2 - 130 y + 1 = 0$$ 6. **Use the quadratic formula:** $$y = \frac{130 \pm \sqrt{(-130)^2 - 4 \cdot 625 \cdot 1}}{2 \cdot 625} = \frac{130 \pm \sqrt{16900 - 2500}}{1250} = \frac{130 \pm \sqrt{14400}}{1250}$$ 7. **Calculate the square root:** $$\sqrt{14400} = 120$$ 8. **Find the two possible values for $y$:** $$y_1 = \frac{130 + 120}{1250} = \frac{250}{1250} = \frac{\cancel{250}}{\cancel{1250}} = \frac{1}{5}$$ $$y_2 = \frac{130 - 120}{1250} = \frac{10}{1250} = \frac{\cancel{10}}{\cancel{1250}} = \frac{1}{125}$$ 9. **Recall $y = 5^x$, solve for $x$:** $$5^x = \frac{1}{5} = 5^{-1} \implies x = -1$$ $$5^x = \frac{1}{125} = 5^{-3} \implies x = -3$$ **Final answer:** $$x = -1 \text{ or } x = -3$$