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Prove that x^(logy-logz)*y^(logz-logx)*z^(logx-logy) = 1​

User Jeffre
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Answer:

We can start by using logarithmic properties. If a, b, and c are positive numbers, then:

loga^b = b * loga

Using this property, we can rewrite the expression as:

x^(logy - logz) * y^(logz - logx) * z^(logx - logy) = x^(logy/logx - logz/logx) * y^(logz/logy - logx/logy) * z^(logx/logz - logy/logz)

Using the change-of-base formula, we can simplify the exponents:

x^(logy/logx - logz/logx) = (logy/logx) / (logz/logx) = logy/logz

Similarly,

y^(logz/logy - logx/logy) = logz/logx

and

z^(logx/logz - logy/logz) = logx/logy

Now, we can substitute these results back into the original expression:

x^(logy - logz) * y^(logz - logx) * z^(logx - logy) = logy/logz * logz/logx * logx/logy

By the transitive property of logarithms, we have:

logy/logz * logz/logx * logx/logy = logy/logx

Finally, using the change-of-base formula again, we can simplify this result:

logy/logx = y^(1/logx) / x^(1/logx)

Since y and x are positive numbers, their exponents must also be positive. Therefore, we have:

y^(1/logx) / x^(1/logx) = 1

So,

x^(logy - logz) * y^(logz - logx) * z^(logx - logy) = 1

And we have proven that the expression is equal to 1.

Explanation:

User JChristian
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