To prove that

$\frac{(\log^3 + \log^2)(\log^3 - \log^2)}{2} = \frac{1}{2} \log \frac{6}{\sqrt{3}}$

we'll apply the product law and quotient law of logarithms.

Step 1: Simplify the Left Side

Consider the expression on the left: $\frac{(\log^3 + \log^2)(\log^3 - \log^2)}{2}.$

Notice that $(\log^3 + \log^2)(\log^3 - \log^2)$ is in the form $(a + b)(a - b) = a^2 - b^2$. Applying this identity, we get:

$(\log^3 + \log^2)(\log^3 - \log^2) = (\log^3)^2 - (\log^2)^2.$

This simplifies to:

$(\log^3)^2 - (\log^2)^2 = (\log 3)^2 - (\log 2)^2.$

Thus, the left side becomes:

$\frac{(\log 3)^2 - (\log 2)^2}{2}.$

Step 2: Apply the Difference of Squares and Logarithmic Identities

The expression $(\log 3)^2 - (\log 2)^2$ can be rewritten as:

$\frac{(\log 3)^2 - (\log 2)^2}{2} = \frac{1}{2} \left( \log 3 + \log 2 \right) \left( \log 3 - \log 2 \right).$

Using the product law of logarithms, $\log 3 + \log 2 = \log (3 \cdot 2) = \log 6$, and using the quotient law of logarithms, $\log 3 - \log 2 = \log \frac{3}{2}$. So, we get:

$\frac{1}{2} \left( \log 6 \cdot \log \frac{3}{2} \right).$

This simplifies to:

$\frac{1}{2} \log \frac{6}{\sqrt{3}},$

which matches the right side of the original expression.

Conclusion

Thus, we have shown that

$\frac{(\log^3 + \log^2)(\log^3 - \log^2)}{2} = \frac{1}{2} \log \frac{6}{\sqrt{3}}.$

Would you like more details on any step, or have any questions?

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Related Questions

1. How would the proof change if different values were substituted for $\log 3$ and $\log 2$?
2. Can you derive similar identities for $\log$ terms raised to different powers?
3. How do product and quotient laws of logarithms simplify complex logarithmic expressions?
4. What other identities can be proven using difference of squares with logarithmic terms?
5. How can logarithmic identities assist in simplifying exponential equations?

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Tip

When simplifying logarithmic expressions, look for patterns like sums or differences that can convert into a single log term using product and quotient rules.