Find an expression for x, in terms of $e$, for which $\ln(3 - x) = \ln x + 3$. Solve the equation $\log_{5} x + \log_{25} x = 4$.

1. Express the constant as a logarithm and combine the terms on the right-hand side.

$\ln(3 - x) = \ln x + \ln(e^3)$
$\ln(3 - x) = \ln(x e^3)$

2. Equate the arguments of the logarithms.

$3 - x = x e^3$

3. Rearrange the equation to make $x$ the subject.

$3 = x e^3 + x$
$3 = x(e^3 + 1)$
$x = \frac{3}{e^3 + 1}$

1. Change the base of the second logarithm to base 5

$\log_{5} x + \frac{\log_{5} x}{\log_{5} 25} = 4$

2. Simplify the equation

$\log_{5} 25 = 2$
$\log_{5} x + \frac{\log_{5} x}{2} = 4$
$\frac{3}{2} \log_{5} x = 4$

3. Solve for x

$\log_{5} x = \frac{8}{3}$
$x = 5^{\frac{8}{3}}$