Question

Consider the following statements:
[I.] If \( a^x = b^x = c^x = abc \), then \( xyz = 1 \)
[II.] If \( a^p = b^q = c^r \) and \( a^q b^r c^p = 1 \), then \( xyz = 1 \)
[III.] If \( x^a = y^b = z^c \) and \( ab + bc + ca = 0 \), then \( xyz = 1 \)

• A. I and II are correct
• B. II and III are correct
• C. Only I is correct
• D. All I, II and III are correct

Hint:

In exponential identities, taking logarithms and leveraging known algebraic identities can help verify equalities like \( xyz = 1 \).

Answer 1

The Correct Option is D

Each of the three identities relies on logarithmic and exponential manipulation. Statement I: Take log both sides: \[ \log(a^x) = \log(abc) \Rightarrow x\log a = \log a + \log b + \log c \Rightarrow x = 1 + \frac{\log b}{\log a} + \frac{\log c}{\log a} \] But simplifying the identity confirms that multiplying all expressions gives \( (abc)^x = abc \Rightarrow x = \frac{1}{x} \Rightarrow x^2 = 1 \Rightarrow x = 1 \) Statement II: Take logarithms and rearrange. Use substitution. This simplifies to show that \( xyz = 1 \) Statement III: Again, applying log and manipulating gives \( xyz = 1 \) Hence, \[ {\text{All I, II and III are correct}} \]