I have two equations:
$a_1 = \sqrt{b_1}$ and $a_2 = \sqrt{b_2}$
I divde these
$\frac{a_1}{a_2} = \frac{\sqrt{b_1}}{\sqrt{b_2}}$
and solve for $a_1$
$a_1 = a_2 \frac{\sqrt{b_1}}{\sqrt{b_2}}$
Fine. Now I take the same initial equations, but logarithmise these
$Ln(a_1) = Ln(\sqrt{b_1})$ and $Ln(a_2) = Ln(\sqrt{b_2})$
Again divde these
$\frac{Ln(a_1)}{Ln(a_2)} = \frac{0.5 \, Ln(b_1)}{0.5 \, Ln(b_2)}$
and solve for $Ln(a_1)$
$Ln(a_1) = Ln(a_2) \frac{Ln(b_1)}{Ln(b_2)}$
Applying logarithm laws results in
$a_1 = a_2^{\frac{Ln(b_1)}{Ln(b_2)}}$
If this is correct, it is sort of confusing that $a_1$ can expressed in a linear, but also in an exponential form. Or is it simply that in the second case the initial logarithmisation is delogarithmised by the exponential function? Interestingly, also, the square root disappears in the logarithmised form.
$$ a_2^{\ln(b_1)/\ln(b_2)}=e^{\frac{\ln(b_1)}{\ln(b_2)}\ln(a_2)} $$ Using $\displaystyle \ln(a_2)=\frac{\ln(b_2)}{2}$ you have
In fact, it is not surprising because you wrote $$ \frac{a_1}{a_2}=\sqrt{\frac{b_1}{b_2}} $$ So taking the log here gives you $$ \ln\left(\frac{a_1}{a_2}\right)=\frac{1}{2}\ln\left(\frac{b_1}{b_2}\right)=\frac{1}{2}\ln\left(b_1\right)-\frac{1}{2}\ln(b_2)=\ln(a_1)-\ln(a_2) $$