Logarithm Equation Simplification

Begin with the given equation:

$ \log (p^2 + q^2) = \log p + \log q + 2 \log 3 $

Use logarithm properties on the right side:

• $n \log a = \log(a^n) \implies 2 \log 3 = \log(3^2) = \log 9$.
• $\log a + \log b = \log(ab) \implies \log p + \log q = \log(pq)$.

The equation simplifies to:

$ \log (p^2 + q^2) = \log (pq) + \log 9 $

$ \log (p^2 + q^2) = \log (9pq) $

Equating Logarithm Arguments

Since $\log x = \log y$ implies $x=y$ for positive $x, y$, we set the arguments equal:

$ p^2 + q^2 = 9pq $

Deriving the Ratio Value

Divide the equation $p^2 + q^2 = 9pq$ by $pq$ (since $p, q$ are non-zero):

$ \frac{p^2}{pq} + \frac{q^2}{pq} = \frac{9pq}{pq} $

$ \frac{p}{q} + \frac{q}{p} = 9 $

We need to find the value of $\frac{p^4+q^4}{p^2q^2}$. Rewrite this expression:

$ \frac{p^4+q^4}{p^2q^2} = \frac{p^4}{p^2q^2} + \frac{q^4}{p^2q^2} = \frac{p^2}{q^2} + \frac{q^2}{p^2} $

Consider the square of the ratio sum:

$ \left(\frac{p}{q} + \frac{q}{p}\right)^2 = \left(\frac{p}{q}\right)^2 + \left(\frac{q}{p}\right)^2 + 2 \left(\frac{p}{q}\right) \left(\frac{q}{p}\right) $

$ \left(\frac{p}{q} + \frac{q}{p}\right)^2 = \frac{p^2}{q^2} + \frac{q^2}{p^2} + 2 $

Substitute the value $9$ for $\left(\frac{p}{q} + \frac{q}{p}\right)$:

$ 9^2 = \frac{p^2}{q^2} + \frac{q^2}{p^2} + 2 $

$ 81 = \frac{p^2}{q^2} + \frac{q^2}{p^2} + 2 $

Isolate the term $\frac{p^2}{q^2} + \frac{q^2}{p^2}$:

$ \frac{p^2}{q^2} + \frac{q^2}{p^2} = 81 - 2 = 79 $

Thus, the value of $\frac{p^4+q^4}{p^2q^2}$ is 79.