Question

Consider two distinct positive real numbers \(m, n\), with \(m > n\).

Let \(x = n^{\log_{10}(m)}\) and \(y = m^{\log_{10}(n)}\). The relation between \(x\) and \(y\) is _______.

• A. \(x > y\)
• B. \(x < y\)
• C. \(x = y\)
• D. \(x = \log_{10}(y)\)

Hint:

Take \(\log_{10}\) of both \(x\) and \(y\) and use the rule \(\log_{10}(a^k) = k \log_{10}(a)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question.
We are given two distinct positive real numbers \(m\) and \(n\), with \(m > n\), and two quantities built from them, \(x = n^{\log_{10}(m)}\) and \(y = m^{\log_{10}(n)}\). We need to find how \(x\) and \(y\) compare. Since \(m\) and \(n\) are both positive, \(x\) and \(y\) are also both positive, so it is safe to take \(\log_{10}\) of both sides to compare them.

Step 2: Take the logarithm of x.
Using the log power rule \(\log_{10}(a^{k}) = k \cdot \log_{10}(a)\), with \(a = n\) and \(k = \log_{10}(m)\):
\[ \log_{10}(x) = \log_{10}\left(n^{\log_{10}(m)}\right) = \log_{10}(m) \cdot \log_{10}(n) \]

Step 3: Take the logarithm of y.
The same power rule applies here, with \(a = m\) and \(k = \log_{10}(n)\):
\[ \log_{10}(y) = \log_{10}\left(m^{\log_{10}(n)}\right) = \log_{10}(n) \cdot \log_{10}(m) \]

Step 4: Compare the two results.
\(\log_{10}(x)\) and \(\log_{10}(y)\) are both equal to the same product, \(\log_{10}(m) \cdot \log_{10}(n)\), just written in a different order. Multiplication does not care about order, so:
\[ \log_{10}(x) = \log_{10}(y) \]
Since \(\log_{10}\) is a one to one function on positive numbers (equal logs mean equal arguments), this gives \(x = y\). Notice this holds true no matter what specific positive values \(m\) and \(n\) take, and the condition \(m > n\) turns out not to matter at all for this equality.

Step 5: Why the other options are wrong.
Options (A) and (B) claim a strict inequality between \(x\) and \(y\), but we have shown their logarithms are exactly equal for any valid \(m, n\), so neither can be strictly greater. Option (D) claims \(x = \log_{10}(y)\), but since \(x = y\) here, this would require \(y = \log_{10}(y)\), which is false for a generic positive \(y\) (for example \(y=10\) gives \(\log_{10}(10)=1\), not 10).

Final Answer:
\(x\) and \(y\) are always equal. \[ \boxed{x = y} \]