If $$\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-B$$ and $$\log_2{a}=\frac{1}{3}$$, then $$\log_3{a}$$ equals
$$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$$...........(1)
$$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$$.............(2)
and finally $$\log_a2=3$$
Substituting this in (1) we get $$\log_a5+\log_a3=A-3$$
Now we have two equations in two variables (1) and (2) . On solving we get
$$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$$
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