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 compare, or use the identity \(a^{\log_{10}(b)} = b^{\log_{10}(a)}\).

Answer 1

The Correct Option is C

Step 1: Write down what is given.
We have two distinct positive real numbers \(m\) and \(n\), with \(m > n > 0\).
We define \(x = n^{\log_{10}(m)}\) and \(y = m^{\log_{10}(n)}\), and we must compare \(x\) and \(y\).

Step 2: Take logarithms of both sides.
Since \(x\) and \(y\) are both positive (a positive base raised to any real power stays positive), we can safely take \(\log_{10}\) of both without changing the direction of any equality.
\[ \log_{10}(x) = \log_{10}\left(n^{\log_{10}(m)}\right) \]
Use the power rule of logarithms, \(\log_{10}(a^{b}) = b \cdot \log_{10}(a)\), with \(a = n\) and \(b = \log_{10}(m)\):
\[ \log_{10}(x) = \log_{10}(m) \cdot \log_{10}(n) \]

Step 3: Do the same for \(y\).
\[ \log_{10}(y) = \log_{10}\left(m^{\log_{10}(n)}\right) = \log_{10}(n) \cdot \log_{10}(m) \]

Step 4: Compare the two results.
\[ \log_{10}(x) = \log_{10}(m)\cdot\log_{10}(n) = \log_{10}(y) \]
Both expressions are exactly the same product, \(\log_{10}(m) \cdot \log_{10}(n)\), so \(\log_{10}(x)\) equals \(\log_{10}(y)\). Since \(\log_{10}\) is a one-to-one (strictly increasing) function, equal logarithms mean the original numbers are equal too:
\[ x = y \]
This holds for any valid choice of distinct positive \(m > n\); for example \(m = 100\), \(n = 10\) gives \(x = 10^{\log_{10}100} = 10^{2} = 100\) and \(y = 100^{\log_{10}10} = 100^{1} = 100\), confirming \(x = y\).

Step 5: Why the other options fail.
Options (A) and (B) claim a strict inequality, but the algebra shows the two sides are always exactly equal, regardless of which specific \(m > n\) is chosen, so no strict inequality can hold in general.
Option (D) mixes up the relation; \(x\) equals \(y\) itself, not \(\log_{10}(y)\).

Step 6: Final Answer.
\[ \boxed{x = y} \]