1. **State the problem:** Solve the equation $$\ln\left(\frac{e^3}{\sqrt{x+0}}\right) \cdot x = \ln\left(\frac{e^3}{\sqrt{n+1}}\right)$$ for $x$. 2. **Rewrite the logarithmic expressions:** Recall that $$\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$$ and $$\sqrt{x} = (x)^{1/2}$$. 3. Apply these rules: $$\ln\left(\frac{e^3}{\sqrt{x}}\right) = \ln(e^3) - \ln(x^{1/2}) = 3 - \frac{1}{2} \ln(x)$$ Similarly, $$\ln\left(\frac{e^3}{\sqrt{n+1}}\right) = 3 - \frac{1}{2} \ln(n+1)$$ 4. Substitute back into the equation: $$\left(3 - \frac{1}{2} \ln(x)\right) \cdot x = 3 - \frac{1}{2} \ln(n+1)$$ 5. Distribute $x$: $$3x - \frac{x}{2} \ln(x) = 3 - \frac{1}{2} \ln(n+1)$$ 6. This is a transcendental equation in $x$ and generally cannot be solved algebraically for $x$ in closed form. 7. If $n$ is known, numerical methods (like Newton-Raphson) can approximate $x$. --- **Additional problem:** Given $$p(x) = 3e^x - 2e^{-x}$$ - Calculate $p(5)$: $$p(5) = 3e^5 - 2e^{-5} = 3e^5 - \frac{2}{e^5}$$ - Find $x$ such that $p(x) = 0$: $$3e^x - 2e^{-x} = 0$$ $$3e^x = 2e^{-x}$$ $$3e^{2x} = 2$$ $$e^{2x} = \frac{2}{3}$$ $$2x = \ln\left(\frac{2}{3}\right)$$ $$x = \frac{1}{2} \ln\left(\frac{2}{3}\right)$$ **Final answers:** - The original equation requires numerical methods for $x$ given $n$. - $p(5) = 3e^5 - \frac{2}{e^5}$ - $p(x) = 0$ at $$x = \frac{1}{2} \ln\left(\frac{2}{3}\right)$$