High School

If $x^y < 15$, then the expression in Column I is true for which conditions in Column II?

| Column-I | Column-II |
|---|---|
| I) $\frac{1}{log_x y} + \frac{1}{log_y x} > y$ | P) $log_{15}(\frac{15}{y}) < \frac{5}{26}, e^x < \frac{225}{y}$ |
| II) $ln x < \frac{4}{5}$ | Q) $log_5(\frac{x}{y}) > \frac{1}{6}, y^5 < 3125$ |
| III) $log_{15}(x) < \frac{5}{7}$ | R) For all $x, y$ |
| IV) $y \leq \frac{5}{7} log_5(15)$ | S) $log_{27}(\frac{27}{y}) > \frac{8y}{15}, 3^y < y$ |

A I$\rightarrow$R, II$\rightarrow$S, III$\rightarrow$P, IV$\rightarrow$Q

B I$\rightarrow$Q, II$\rightarrow$P, III$\rightarrow$S, IV$\rightarrow$R

C I$\rightarrow$Q, II$\rightarrow$S, III$\rightarrow$P, IV$\rightarrow$R

D I$\rightarrow$R, II$\rightarrow$P, III$\rightarrow$S, IV$\rightarrow$Q

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].

  1. 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.

  2. 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.

  3. 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.

  4. 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.