Solve the equations.

1. Express the constant term as a logarithm to base 2.

$\log_2(x+4) = 2\log_2 x - \log_2 2$

2. Apply the power law and quotient law of logarithms to combine the right-hand side.

$\log_2(x+4) = \log_2 x^2 - \log_2 2$
$\log_2(x+4) = \log_2\left(\frac{x^2}{2}\right)$

3. Equate the arguments of the logarithms and form a quadratic equation.

$x + 4 = \frac{x^2}{2}$
$2x + 8 = x^2$
$x^2 - 2x - 8 = 0$

4. Solve the quadratic equation and reject any invalid roots.

$(x - 4)(x + 2) = 0$
$x = 4$ or $x = -2$
Since $x > 0$ for $\log_2 x$ to be defined, we reject $x = -2$.
Therefore, $x = 4$.

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

Using the change of base formula, $\log_y 5 = \frac{1}{\log_5 y}$.
The equation becomes:
$10\left(\frac{1}{\log_5 y}\right) + 3 = \log_5 y$

2. Use substitution to form a quadratic equation.

Let $u = \log_5 y$. The equation becomes:
$\frac{10}{u} + 3 = u$
Multiply throughout by $u$:
$10 + 3u = u^2$
$u^2 - 3u - 10 = 0$

3. Solve the quadratic equation for $u$.

$(u - 5)(u + 2) = 0$
$u = 5 \quad \text{or} \quad u = -2$

4. Solve for $y$ using the definition of logarithms.

Substitute back $u = \log_5 y$:
If $\log_5 y = 5$, then $y = 5^5 = 3125$.
If $\log_5 y = -2$, then $y = 5^{-2} = \frac{1}{25}$.