Solve the recurrence relation by taking the logarithm of both sides and making the substitution ${b}_{n}=\mathrm{lg}{a}_{n}$

Solve this recurrence relation:

${a}_{n}={\left(\frac{{a}_{n-2}}{{a}_{n-1}}\right)}^{\frac{1}{2}}$

by taking the logarithm of both sides and making the substitution

${b}_{n}=\mathrm{lg}{a}_{n}$

A couple years ago I took precalc and a couple years before that I took College Algebra to brush up since before then I'd been out of school for a couple of years already. So I'm extremely rusty and this first step, getting the logarithm of both sides, has me confused enough... Like how would I even get the logarithm of the left side?? and then what exactly is that "substitution" supposed to be substituting?

So log'ing both sides I get

$\mathrm{ln}({a}_{n})=\frac{1}{2}\mathrm{ln}({a}_{n-2})-\frac{1}{2}\mathrm{ln}({a}_{n-1})$