#### If 2^{x} = 3^{y} = 6^{-z}, prove that 1/x + 1/y + 1/z = 0.

**Solution:**

Consider

2^{x} = 3^{y} = 6^{-z} = k

Here

2^{x} = k

We can write it as

2 = (k)^{1/x}

3^{y} = k

We can write it as

3 = (k)^{1/y}

6^{-z} = k

We can write it as

6 = (k)^{-1/z}

So we get

2 × 3 = 6

(k)^{1/x} × (k)^{1/y} = (k)^{-1/z}

By further calculation

(k)^{1/x + 1/y} = (k)^{-1/z}

We get

1/x + 1/y = – 1/z

1/x + 1/y + 1/z = 0

Therefore, it is proved.

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