通过使用合适的代换，或用其他方法，解方程 $3\log_6 x = \log_x 3 + \log_x 2 + \log_5 25.$ 在给定的坐标系中画出 $y = \log_6 x$ 的函数图象草图。标出与坐标轴的所有交点。 已知一条曲线的方程为 $y = e^{2x+1} + 2e^{x+1} - 3e.$ 说明为什么该曲线没有驻点。 求该曲线 $y = e^{2x+1} + 2e^{x+1} - 3e.$ 的 x 轴截距。

1. Simplify the terms on the right hand side of the equation.

$\log_x 3 + \log_x 2 = \log_x (3 \times 2) = \log_x 6$
$\log_5 25 = \log_5 (5^2) = 2$
The equation becomes: $3\log_6 x = \log_x 6 + 2$

2. Apply the change of base formula to express all logarithms in the same base.

$\log_x 6 = \frac{1}{\log_6 x}$
So, $3\log_6 x = \frac{1}{\log_6 x} + 2$

3. Use a substitution to form a quadratic equation.

Let $u = \log_6 x$.
$3u = \frac{1}{u} + 2$
$3u^2 - 2u - 1 = 0$

4. Solve the quadratic equation and find the corresponding values of $x$.

$(3u + 1)(u - 1) = 0$
$u = -\frac{1}{3}$ or $u = 1$

When $u = 1$:
$\log_6 x = 1 \implies x = 6^1 = 6$

When $u = -\frac{1}{3}$:
$\log_6 x = -\frac{1}{3} \implies x = 6^{-\frac{1}{3}}$

1. Identify the general shape and asymptote of the logarithmic graph

The graph of $y = \log_6 x$ is a standard logarithmic curve. It has a vertical asymptote at $x = 0$ (the $y$-axis) and is strictly increasing for $x > 0$.

2. Determine the axial intercept

Set $y = 0$ to find the $x$-intercept:
$\log_6 x = 0 \implies x = 6^0 = 1$.
The $x$-intercept is $(1, 0)$. There is no $y$-intercept since $x > 0$.

1. Find the first derivative of the curve with respect to $x$.

$\frac{dy}{dx} = 2e^{2x+1} + 2e^{x+1}$

2. State the condition for a stationary point and analyze the terms of the derivative.

For a stationary point, $\frac{dy}{dx} = 0$.
For all real values of $x$, $e^{2x+1} > 0$ and $e^{x+1} > 0$.

3. Conclude that the derivative cannot be zero.

Therefore, $\frac{dy}{dx} = 2e^{2x+1} + 2e^{x+1} > 0$ for all real $x$.
Since $\frac{dy}{dx}$ can never be equal to $0$, the curve has no stationary point.

1. Set $y = 0$ to find the x-intercept and simplify the equation

$e^{2x+1} + 2e^{x+1} - 3e = 0$
$e(e^{2x}) + 2e(e^x) - 3e = 0$
Dividing by $e$ (since $e \neq 0$):
$e^{2x} + 2e^x - 3 = 0$

2. Solve the resulting quadratic equation in $e^x$

Let $u = e^x$. The equation becomes:
$u^2 + 2u - 3 = 0$
$(u + 3)(u - 1) = 0$
$u = -3 \quad \text{or} \quad u = 1$
$e^x = -3 \quad \text{or} \quad e^x = 1$

3. Solve for $x$ and reject any invalid solutions

Since $e^x > 0$ for all real $x$, $e^x = -3$ has no real solution.
For $e^x = 1$:
$x = \ln 1$
$x = 0$