Answer :
To solve this problem, we need to match the conditions in Column I with those in Column II based on the restrictions given by the inequality [tex]x^y < 15[/tex].
Expression I: [tex]\frac{1}{\log_x y} + \frac{1}{\log_y x} > y[/tex]
- This is always true for all values of [tex]x[/tex] and [tex]y[/tex] that satisfy the condition [tex]x^y < 15[/tex]. This is because [tex]\frac{1}{\log_x y} + \frac{1}{\log_y x} = 2[/tex] by the properties of logarithms (because it equals [tex]\log_y x + \log_x y = 2[/tex]). Since [tex]2[/tex] is a constant greater than any [tex]y[/tex] satisfying [tex]x^y < 15[/tex], it holds always in this domain.
Expression II: [tex]\ln x < \frac{4}{5}[/tex]
- The condition that matches here is when [tex]\log_{27}(\frac{27}{y}) > \frac{8y}{15} \text{ and } 3^y < y[/tex]. Here we are comparing logarithms and exponential functions based on different bases, leading to a very specific inequality.
Expression III: [tex]\log_{15}(x) < \frac{5}{7}[/tex]
- The correct condition for this is [tex]\log_{15}(\frac{15}{y}) < \frac{5}{26}, e^x < \frac{225}{y}[/tex], which involves expressions that relate [tex]x[/tex] and [tex]y[/tex] according to the given boundaries.
Expression IV: [tex]y \leq \frac{5}{7} \log_5(15)[/tex]
- This corresponds to [tex]\log_5(\frac{x}{y}) > \frac{1}{6}, y^5 < 3125[/tex] since solving this involves comparison of logarithmic bases and inequalities.
Putting it together, we have:
- I corresponds to R: For all [tex]x, y[/tex]
- II corresponds to S: [tex]\log_{27}(\frac{27}{y}) > \frac{8y}{15}, 3^y < y[/tex]
- III corresponds to P: [tex]\log_{15}(\frac{15}{y}) < \frac{5}{26}, e^x < \frac{225}{y}[/tex]
- IV corresponds to Q: [tex]\log_5(\frac{x}{y}) > \frac{1}{6}, y^5 < 3125[/tex]
Therefore, the correct matching answer is option A.