Question - Algebraic Evaluation of an Expression with Exponents

\(6^z = (2 \cdot 3)^z = 2^z \cdot 3^z\)

Since \(2^x = 3^y = 6^z\), we can say \(2^x = 2^z \cdot 3^z\).

Therefore, \(x = z \cdot (\log_2{2} + \log_2{3}) = z + z \cdot \log_2{3}\).

Similarly, \(3^y = 2^z \cdot 3^z\) implies \(y = z \cdot (\log_3{2} + \log_3{3}) = z \cdot \log_3{2} + z\).

We find \(\log_2{3}\) and \(\log_3{2}\) by changing the base:

\(\log_2{3} = \frac{1}{\log_3{2}}\)

\(x = z + z \cdot \log_2{3} = z + \frac{z}{\log_3{2}}\)

\(y = z \cdot \log_3{2} + z\)

We then calculate the sum:

\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{z + \frac{z}{\log_3{2}}} + \frac{1}{z \cdot \log_3{2} + z} + \frac{1}{z}\)

By finding a common denominator, we have:

\(\frac{\log_3{2}(\log_3{2} + 1) + 1 + \log_3{2}(\log_3{2} + 1)}{z(\log_3{2} + 1)}\)

This simplifies to:

\(\frac{2\log_3{2}(\log_3{2} + 1) + 1}{z(\log_3{2} + 1)}\)

Since \(\log_3{2}(\log_3{2} + 1) = 1\), we get:

\(\frac{2 \cdot 1 + 1}{z(\log_3{2} + 1)} = \frac{3}{z(\log_3{2} + 1)}\)

Now, \(y = z \cdot \log_3{2} + z\) gives us \(y = z(\log_3{2} + 1)\), hence substituting \(y\) into the denominator yields:

\(\frac{3}{y}\)

Since \(2^x=3^y\), it follows that \(x=y\), thus \(\frac{3}{y} = \frac{3}{x}\).

Finally, \(2^x=3^y\) implies \(1 = \frac{3}{x}\), so the value of the sum \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) is 3.