We are given a sequence $(u_n)$ defined by $u_0 = 9e$ and $u_{n+1} = 3\sqrt{u_n}$ for all $n \in \mathbb{N}$. We are also given $v_n = \ln\left(\frac{u_n}{9}\right)$ for all $n \in \mathbb{N}$. The first question is to calculate $u_1$ and $v_0$. The second question is to show that the sequence $(v_n)$ is a geometric sequence and to find its first term and common ratio.

1. Problem Description

We are given a sequence $(u_n)$ defined by $u_0 = 9e$ and $u_{n+1} = 3\sqrt{u_n}$ for all $n \in \mathbb{N}$. We are also given $v_n = \ln\left(\frac{u_n}{9}\right)$ for all $n \in \mathbb{N}$. The first question is to calculate $u_1$ and $v_0$. The second question is to show that the sequence $(v_n)$ is a geometric sequence and to find its first term and common ratio.

2. Solution Steps

First, we calculate $u_1$:

$u_1 = 3\sqrt{u_0} = 3\sqrt{9e} = 3 \cdot 3\sqrt{e} = 9\sqrt{e}$.

Next, we calculate $v_0$:

$v_0 = \ln\left(\frac{u_0}{9}\right) = \ln\left(\frac{9e}{9}\right) = \ln(e) = 1$.

Now, we need to show that $(v_n)$ is a geometric sequence. We compute $v_{n+1}$ in terms of $v_n$:

$v_{n+1} = \ln\left(\frac{u_{n+1}}{9}\right) = \ln\left(\frac{3\sqrt{u_n}}{9}\right) = \ln\left(\frac{\sqrt{u_n}}{3}\right) = \ln(\sqrt{u_n}) - \ln(3) = \frac{1}{2}\ln(u_n) - \ln(3)$.

Since $v_n = \ln\left(\frac{u_n}{9}\right) = \ln(u_n) - \ln(9)$, we have $\ln(u_n) = v_n + \ln(9) = v_n + 2\ln(3)$.

Substituting this into the expression for $v_{n+1}$:

$v_{n+1} = \frac{1}{2}(v_n + 2\ln(3)) - \ln(3) = \frac{1}{2}v_n + \ln(3) - \ln(3) = \frac{1}{2}v_n$.

Thus, $v_{n+1} = \frac{1}{2} v_n$, which means $(v_n)$ is a geometric sequence with common ratio $q = \frac{1}{2}$ and first term $v_0 = 1$.

3. Final Answer

$u_1 = 9\sqrt{e}$

$v_0 = 1$

$(v_n)$ is a geometric sequence with first term $v_0 = 1$ and common ratio $q = \frac{1}{2}$.