Question

Let 3^a=4, 4^b=5, 5^c=6, 6^d=7, 7^e=8, 8^f=9 then find the value od the product abcdef.

Answer 1

Given that 3^a=4, 4^b=5, 5^c=6, 6^d=7, 7^e=8, 8^f=9
since 3^a=4 ------(1)
and 4^b=5. ------(2)
from equation (1) we can write 4 as 3^a
(3^a)^b = 5
3^(ab) = 5
similarly,
(3^ab)^c = 6
3^(abc) = 6

(3^abc)^d = 7
3^(abcd) = 7

(3^abcd)^e = 8
3^(abcde) = 8

(3^abcde)^f = 9
3^(abcdef) = 3^2
since bases are same so we can compare the exponents
abcdef = 2
Hope it helps!!!!!

Answer 2

Answer:

The product of abcdef = 2

Step-by-step explanation:

Given that $3^{a} =4$, $4^{b} =5$, $5^{c} =6$, $6^{d} =7$, $7^{e} =8$ , $8^{f} = 9$

since $3^{a} =4$ ...........(1)

and   $4^{b} =5$ ...........(2)

∴  $(3^a)^b = 5$

   → $3^{ab} = 5$

similarly,

             $(3^{ab)^c = 6$

          → $3^{abc} = 6$

Again,

             $(3^{abc)^d = 7$

          → $3^{abcd} = 7$

          →  $(3^{abcd})^e = 8$

           ∴  $3^{abcde} = 8$

   Finally, $(3^{abcde})^f = 9$

             → $3^{abcde(f) = 3^{2}$

since bases are same, so we can compare the exponents and finally we will get the product of abcdef = 2.