To prove that $z = \frac{xy}{x+y}$ given that $3^x = 4^y = 12^z$, we can follow these steps:

Step 1: Express the equality in logarithmic form

Since $3^x = 4^y = 12^z$, let's denote this common value as $k$. Therefore, we have: $k = 3^x = 4^y = 12^z$ Taking the natural logarithm (or logarithm to any base) of all sides, we get: $\ln(k) = x \ln(3) = y \ln(4) = z \ln(12)$

Step 2: Express $x$, $y$, and $z$ in terms of $\ln(k)$

From the logarithmic expressions, we have: $x = \frac{\ln(k)}{\ln(3)}, \quad y = \frac{\ln(k)}{\ln(4)}, \quad z = \frac{\ln(k)}{\ln(12)}$

Step 3: Use the relationship between $\ln(12)$, $\ln(3)$, and $\ln(4)$

Recall that $12 = 3 \times 4$, so: $\ln(12) = \ln(3 \times 4) = \ln(3) + \ln(4)$ Thus, we can express $z$ as: $z = \frac{\ln(k)}{\ln(12)} = \frac{\ln(k)}{\ln(3) + \ln(4)}$

Step 4: Express $z$ in terms of $x$ and $y$

Now, substitute the expressions for $x$ and $y$ from Step 2: $z = \frac{\ln(k)}{\ln(3) + \ln(4)} = \frac{\frac{\ln(k)}{\ln(3)} \cdot \frac{\ln(k)}{\ln(4)}}{\frac{\ln(k)}{\ln(3)} + \frac{\ln(k)}{\ln(4)}}$ Simplifying further: $z = \frac{xy}{x + y}$ Thus, we have proven that: $z = \frac{xy}{x + y}$

Conclusion

This completes the proof that given $3^x = 4^y = 12^z$, we indeed have $z = \frac{xy}{x + y}$.

Would you like more details or have any other questions?

Here are some related questions:

1. How can this method be applied to similar exponential equations?
2. What would change if $3^x$ and $4^y$ were raised to different powers?
3. Can this proof be extended to cases where more than three bases are involved?
4. How would the proof differ if we used a different logarithm base?
5. What does this relationship say about the symmetry between $x$, $y$, and $z$?

Tip: Whenever you're working with exponential equations, logarithms are powerful tools to bring the exponents down and analyze the relationships.