If $\log_{12}81 = p$, then $3\frac{4-p}{4+p}$ is equal to

Given: $\log_{12}81 = p$

To find: $3\frac{4-p}{4+p}$

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First, let's recognize that $81 = 3^4$ (since $3 \times 3 \times 3 \times 3 = 81$).

So: $\log_{12}81 = \log_{12}3^4 = p$

Using the power rule of logarithms: $\log_a(x^n) = n\log_a(x)$

$\log_{12}3^4 = 4\log_{12}3 = p$

Therefore: $\log_{12}3 = \frac{p}{4}$

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Now let's work on $3\frac{4-p}{4+p}$.

Since $p = 4\log_{12}3$, we substitute:

$3\frac{4-p}{4+p} = 3\frac{4 - 4\log_{12}3}{4 + 4\log_{12}3}$

Factor out 4 from numerator and denominator:

$= 3\frac{4(1 - \log_{12}3)}{4(1 + \log_{12}3)} = 3\frac{1 - \log_{12}3}{1 + \log_{12}3}$

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Here's a key insight: $\log_{12}12 = 1$ (since any number to the power 1 equals itself).

So we can rewrite:

$1 = \log_{12}12$

$1 - \log_{12}3 = \log_{12}12 - \log_{12}3$

$1 + \log_{12}3 = \log_{12}12 + \log_{12}3$

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Using the quotient rule: $\log_a(x) - \log_a(y) = \log_a(\frac{x}{y})$

And the product rule: $\log_a(x) + \log_a(y) = \log_a(xy)$

$3\frac{\log_{12}12 - \log_{12}3}{\log_{12}12 + \log_{12}3} = 3\frac{\log_{12}(\frac{12}{3})}{\log_{12}(12 \times 3)} = 3\frac{\log_{12}4}{\log_{12}36}$

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The change of base formula states: $\frac{\log_a(x)}{\log_a(y)} = \log_y(x)$

So: $\frac{\log_{12}4}{\log_{12}36} = \log_{36}4$

Therefore: $3\frac{\log_{12}4}{\log_{12}36} = 3\log_{36}4$

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Using $a\log_b(x) = \log_b(x^a)$:

$3\log_{36}4 = \log_{36}4^3 = \log_{36}64$

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Notice that:

$36 = 6^2$

$64 = 8^2$

So: $\log_{36}64 = \log_{6^2}8^2$

Using the property $\log_{a^m}b^n = \frac{n}{m}\log_a(b)$:

$\log_{6^2}8^2 = \frac{2}{2}\log_6(8) = \log_6(8)$

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Therefore: $3\frac{4-p}{4+p} = \log_6(8)$

This problem beautifully demonstrates how multiple logarithm properties work together to simplify complex expressions!