December 27, 2024
Explanation:
(a)
log27 8 ≤ log3 x < 9 1/log2 3
log3^3 8≤ log3 x <9 1/log2 3 [Using logm^n a = 1/n .logm a & 1/logb a = logab]
1/3 .log3 8 ≤log3 x <3^2∙log3 2
log3 (8)^1/3 ≤ log3 x <3^log3 4
log3 2 ≤ log3 x < 4 [Using a^loga x = x]
log3 2 ≤ log3 x < 4∙log3 3
log3 2 ≤ log3 x < log3 (3)^4
log3 2 ≤ log3 x < log3 81
= 2 ≤ x < 81 or [2,81)