Solve for a: Division of Logarithms with Base 7 and 9 Equation

The given problem requires us to solve for $a$ from the equation:

$\frac{4a^2}{\log_7 9} : \log_9 7 = 16$.

First, recognize that the expression $\colon$ represents division, thus:

$\frac{4a^2}{\log_7 9} = \log_9 7 \times 16.$

From the property of logarithms, we know $\log_9 7 = \frac{1}{\log_7 9}$. Hence, we can express the equation as:

$\frac{4a^2}{\log_7 9} = \frac{16}{\log_7 9}.$

By equating both sides and simplifying, we get:

$4a^2 = 16.$

Solving for $a^2$ gives:

$a^2 = 4.$

Taking the square root of both sides, we find:

$a = \pm2.$

Therefore, the value of $a$ is $\pm 2$.