Proof:

Given: 2^x = 3^y = 6^-z

We need to prove that 1/x + 1/y + 1/z = 0.

Solution:

Let's start by expressing each term in terms of the other.

2^x = 3^y

Taking the natural logarithm of both sides, we get:

x ln 2 = y ln 3

x/y = ln 3/ln 2 ...(1)

Similarly,

3^y = 6^-z

Taking the natural logarithm of both sides, we get:

y ln 3 = -z ln 6

y/z = -ln 6/ln 3 ...(2)

We can rewrite the equation (1) as:

1/x = ln 2/ln 3 * 1/y

Similarly, we can rewrite equation (2) as:

1/y = -ln 3/ln 6 * 1/z

Substituting these expressions in 1/x + 1/y + 1/z, we get:

1/x + 1/y + 1/z = ln 2/ln 3 * 1/y + (-ln 3/ln 6) * 1/z + 1/z

Simplifying, we get:

1/x + 1/y + 1/z = (ln 2/ln 3 - ln 3/ln 6) * 1/z

Using the property of logarithms, we can simplify this expression as follows:

1/x + 1/y + 1/z = ln (2/3^ln 6)/ln 3 * 1/z

Now, note that 2/3^ln 6 is a constant, which means that ln (2/3^ln 6) is also a constant. Therefore, as we take the limit of 1/z as z approaches infinity, the entire expression approaches zero.

Hence, we have proved that 1/x + 1/y + 1/z = 0.

Conclusion:

We have shown that if 2^x = 3^y = 6^-z, then 1/x + 1/y + 1/z = 0. This result is important in many areas of mathematics and physics, where logarithmic functions appear frequently.

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