Problem: If x is real and positive and grows beyond all bounds, then log3​(6x−5)−log3​(2x+1) approaches:
Answer Choices:
A. 0
B. 1
C. 3
D. 4
E. no finite number
Solution:
log3​(6x−5)−log3​(2x+1)=log3​2x+16x−5​=log3​2+x1​6−x5​​ for xî€ =0. As x grows beyond all bounds, the last expression approaches log3​26​=log3​3=1.