Answer: A
log₁₀ A = B and log₅ A= C
=> A = 10^B and A = 5^C
Therefore, A = 10^B = 5^C
2^B = 5 ^C-B
2 = 5^(C-B)/B = 5^x  (As given in question)
Thus x = C/B -1

Answer: B
3x² - 7x + 3 = (2x -3)²
x² -5x + 6 =0
=> x= 2 and 3 and in both cases (3x² - 7x + 3) and (2x -3) are positive.
Thus, both the value satisfy the condition. x can be 2 or 3.

Answer: D
(3^{log₂ 9}/9^{log₂ 3}) * (5^1+log₅2/3^2-log₃27) = (3^{log₂ 9}/3^{log₂ 9}) * (5(2)/3^2-3)
= 5(2)(3) = 30.