1. **State the problem:** Solve the equation $$e^{x+1} - 18e^x - 3 = 0$$ for $x$. 2. **Rewrite the equation:** Use the property of exponents: $$e^{x+1} = e^x \cdot e^1 = e \cdot e^x$$. 3. **Substitute:** Let $$y = e^x$$, then the equation becomes: $$e \cdot y - 18y - 3 = 0$$ 4. **Factor the equation:** $$y(e - 18) - 3 = 0$$ 5. **Isolate $y$:** $$y(e - 18) = 3$$ $$y = \frac{3}{e - 18}$$ 6. **Recall substitution:** $$e^x = \frac{3}{e - 18}$$ 7. **Solve for $x$:** Take the natural logarithm on both sides: $$x = \ln\left(\frac{3}{e - 18}\right)$$ 8. **Check domain:** Since $e \approx 2.718$, $e - 18 < 0$, so denominator is negative, making the fraction negative. $e^x$ is always positive, so no real solution exists. **Final answer:** No real solution for $x$. --- **Slug:** exponential equation **Subject:** algebra **Desmos:** {"latex":"e^{x+1} - 18e^x - 3 = 0","features":{"intercepts":true,"extrema":true}} **q_count:** 3