Question:
If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8 and 8^f = 9, then the value of the product abcdef is ?

Solution:

From 3^a = 4  ⇒  a = log_3(4) = ln4 / ln3  
From 4^b = 5  ⇒  b = log_4(5) = ln5 / ln4  
From 5^c = 6  ⇒  c = log_5(6) = ln6 / ln5  
From 6^d = 7  ⇒  d = log_6(7) = ln7 / ln6  
From 7^e = 8  ⇒  e = log_7(8) = ln8 / ln7  
From 8^f = 9  ⇒  f = log_8(9) = ln9 / ln8

Multiply:
abcdef = (ln4/ln3)·(ln5/ln4)·(ln6/ln5)·(ln7/ln6)·(ln8/ln7)·(ln9/ln8)

All intermediate terms cancel telescopically, leaving:
abcdef = ln9 / ln3

But ln9 = ln(3^2) = 2 ln3, so
abcdef = (2 ln3) / ln3 = 2

Answer: 2