Can we find these log expressions in terms of $\log_9 x$?

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By definition, we have that \(x=9^u\).

Using what we know about the logarithm of a product, we get that \[\log_9(3x)=\log_9 3+\log_9 x=\frac{1}{2}+u.\]

Let’s start by writing \(v=\log_x 81\).

By definition, we see that \(81=x^v\) and so using part (a), \(81=(9^u)^v\).

Since \(9^2=81\), we must have that \(uv=2\), or \(v=2/u\). Therefore, \(\log_x 81=2/u\).