If $\log_a 30 = A$, $\log_a(5/3) = - B$ and $\log_2 a = 1 /3$, then $\log_3 a$ equals

We need to find $\log_3 a$ using the given information.

Given:

$\log_a 30 = A$

$\log_a(5/3) = -B$

$\log_2 a = 1/3$

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The easiest approach is to use the change of base relationship:

$\log_3 a = \frac{1}{\log_a 3}$

So we need to find $\log_a 3$.

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Let's add the first two given equations. Since $\log_a(5/3) = -B$, we have:

$A + B = \log_a 30 + \log_a(5/3)$

Using the property $\log_x m + \log_x n = \log_x(mn)$:

$A + B = \log_a \left(30 \times \frac{5}{3}\right) = \log_a \left(\frac{150}{3}\right) = \log_a 50$

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Now let's break down $50$ using prime factors:

$50 = 2 \times 25 = 2 \times 5^2$

So:

$\log_a 50 = \log_a(2 \times 5^2) = \log_a 2 + \log_a 5^2 = \log_a 2 + 2\log_a 5$

Therefore:

$A + B = \log_a 2 + 2\log_a 5$

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From $\log_2 a = \frac{1}{3}$, we can find $\log_a 2$ using the reciprocal relationship:

$\log_a 2 = \frac{1}{\log_2 a} = \frac{1}{\frac{1}{3}} = 3$

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We also need $\log_a 5$. From the first given equation:

$\log_a 30 = A$

Since $30 = 2 \times 3 \times 5$:

$A = \log_a(2 \times 3 \times 5) = \log_a 2 + \log_a 3 + \log_a 5$

Substituting $\log_a 2 = 3$:

$A = 3 + \log_a 3 + \log_a 5$

So:

$\log_a 5 = A - 3 - \log_a 3$

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Substituting back into our equation $A + B = \log_a 2 + 2\log_a 5$:

$A + B = 3 + 2(A - 3 - \log_a 3)$

$A + B = 3 + 2A - 6 - 2\log_a 3$

$A + B = 2A - 3 - 2\log_a 3$

Solving for $\log_a 3$:

$2\log_a 3 = 2A - 3 - A - B$

$2\log_a 3 = A - B - 3$

$\log_a 3 = \frac{A - B - 3}{2}$

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Therefore:

$\log_3 a = \frac{1}{\log_a 3} = \frac{1}{\frac{A - B - 3}{2}} = \frac{2}{A - B - 3}$