If ${\log }_{a}30=A$, ${\log }_a\left(\frac{5}{3}\right)=B$ and ${\log }_a^2=\frac{1}{3}$, then ${\log }_{3}a$ equals

The correct answer is:
$\frac{2}{A-B-3}$

Step-by-step Explanation:

We are given the following logarithmic equations:
1) ${\log }_{a}30=A$
2) ${\log }_a\left(\frac{5}{3}\right)=B$
3) ${\log }_{2}a=\frac{1}{3}$ (which represents the term ${\log }_a^2=\frac{1}{3}$ in the question statement).

Step 1: Simplify the third equation
Using the base-change property of logarithms, we know that ${\log }_{a}b=\frac{1}{{\log }_{b}a}$.
Therefore, we can rewrite the third equation as:
${\log }_{a}2=\frac{1}{{\log }_{2}a}=\frac{1}{1/3}=3$

Step 2: Expand the first two equations using log properties
Using the product rule ${\log }_a\left(xyz\right)={\log }_{a}x+{\log }_{a}y+{\log }_{a}z$ on the first equation:
$A={\log }_{a}30={\log }_a(2·3·5)={\log }_{a}2+{\log }_{a}3+{\log }_{a}5$

Using the quotient rule ${\log }_a\left(\frac{x}{y}\right)={\log }_{a}x-{\log }_{a}y$ on the second equation:
$B={\log }_a\left(\frac{5}{3}\right)={\log }_{a}5-{\log }_{a}3$

Step 3: Relate the expressions for A and B
Let us calculate $A-B$:
$A-B=\left({\log }_{a}2+{\log }_{a}3+{\log }_{a}5\right)-\left({\log }_{a}5-{\log }_{a}3\right)$
Simplifying this yields:
$A-B={\log }_{a}2+2{\log }_{a}3$

Step 4: Substitute the value of ${\log }_{a}2$ and solve for ${\log }_{3}a$
Substituting ${\log }_{a}2=3$ into our simplified equation:
$A-B=3+2{\log }_{a}3$
Rearranging to isolate ${\log }_{a}3$:
$2{\log }_{a}3=A-B-3$
${\log }_{a}3=\frac{A-B-3}{2}$

Finally, we find ${\log }_{3}a$ using the reciprocal property of logarithms:
${\log }_{3}a=\frac{1}{{\log }_{a}3}=\frac{2}{A-B-3}$