不使用计算器，化简 $\log_x 9 \times \log_{27} x$。 解方程 $\ln(2x+e) = 1 + \dfrac{1}{\log_x e}$。

1. Apply the change of base formula to convert both logarithms to a common base (e.g., natural logarithm).

$\log_x 9 \times \log_{27} x = \frac{\ln 9}{\ln x} \times \frac{\ln x}{\ln 27}$

2. Cancel the common term $\ln x$ and express 9 and 27 as powers of 3.

$= \frac{\ln 9}{\ln 27}$
$= \frac{\ln (3^2)}{\ln (3^3)}$

3. Use the power law of logarithms to simplify and obtain the final numerical value.

$= \frac{2\ln 3}{3\ln 3}$
$= \frac{2}{3}$

1. Apply the change of base formula to simplify the fraction

$\frac{1}{\log_x e} = \log_e x = \ln x$

The equation becomes:
$\ln(2x+e) = 1 + \ln x$

2. Express the right hand side as a single logarithm

Since $1 = \ln e$,
$\ln(2x+e) = \ln e + \ln x$
$\ln(2x+e) = \ln(ex)$

3. Equate the arguments of the logarithms

$2x + e = ex$

4. Solve the linear equation for x

$ex - 2x = e$
$x(e - 2) = e$
$x = \frac{e}{e - 2}$