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2^x=3^y=6^-z. What is the value of 1/x+ 1/y+1/z?
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2^x=3^y=6^-z. What is the value of 1/x+ 1/y+1/z?
Given Equation:
2^x = 3^y = 6^-z

Calculating the values:
- Let's start by expressing everything in terms of a common base. We know that 6 = 2 * 3, so we can rewrite the given equation as:
2^x = 3^y = (2 * 3)^-z = 2^-z * 3^-z = (1/2)^z * (1/3)^z
- Now, we can equate the first and second expressions:
2^x = (1/2)^z * (1/3)^z
- Since the bases are the same, we can equate the exponents:
x = -z and y = z

Finding the value of 1/x + 1/y + 1/z:
- We are given x = -z and y = z. Therefore, 1/x + 1/y + 1/z becomes:
1/(-z) + 1/z + 1/z = -1/z + 2/z = 1/z
- So, the value of 1/x + 1/y + 1/z is 1/z.
Therefore, the value of 1/x + 1/y + 1/z is 1/z.
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2^x=3^y=6^-z. What is the value of 1/x+ 1/y+1/z?
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