log_x√6 + log_x√(2/3) = কত?

প্রশ্ন: log_x√6 + log_x√(2/3) = কত?

সমাধান:
log_x√6 + log_x√(2/3)
= log_x6^1/2 + log_x(2/3)^1/2
= (1/2)log_x6 + (1/2) log_x(2/3)
= (1/2)(log_x6 + log_x2/3)
= (1/2)log_x (6 × 2/3)
= (1/2)log_x 4
= (1/2)log_x 2²
= (1/2) × 2log_x2
= log_x 2