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

Answer 1

We are given the following equations:
$3^a = 4 \quad 4^b = 5 \quad 5^c = 6 \quad 6^d = 7 \quad 7^e = 8 \quad 8^f = 9$
We need to find the value of the product $abcdef$.
To solve for each variable:
$3^a = 4 \implies a = \log_3 4$.
$4^b = 5 \implies b = \log_4 5$.
$5^c = 6 \implies c = \log_5 6$.
$6^d = 7 \implies d = \log_6 7$.
$7^e = 8 \implies e = \log_7 8$.
$8^f = 9 \implies f = \log_8 9$.
Now, the value of $abcdef$ is the product of these logarithms:
$abcdef = \log_3 4 \times \log_4 5 \times \log_5 6 \times \log_6 7 \times \log_7 8 \times \log_8 9$
Using the change of base formula for logarithms, we can rewrite each term:
$\log_3 4 = \frac{\log 4}{\log 3}, \quad \log_4 5 = \frac{\log 5}{\log 4}, \quad \log_5 6 = \frac{\log 6}{\log 5}, \dots$
The product simplifies as all the intermediate logarithms cancel out, leaving:
$abcdef = \frac{\log 9}{\log 3} = 2$
Thus, the correct answer is Option (2).

Answer 2

We are given the following equations:
$3^a = 4 \quad 4^b = 5 \quad 5^c = 6 \quad 6^d = 7 \quad 7^e = 8 \quad 8^f = 9$
We need to find the value of the product $abcdef$.
To solve for each variable:
$3^a = 4 \implies a = \log_3 4$.
$4^b = 5 \implies b = \log_4 5$.
$5^c = 6 \implies c = \log_5 6$.
$6^d = 7 \implies d = \log_6 7$.
$7^e = 8 \implies e = \log_7 8$.
$8^f = 9 \implies f = \log_8 9$.
Now, the value of $abcdef$ is the product of these logarithms:
$abcdef = \log_3 4 \times \log_4 5 \times \log_5 6 \times \log_6 7 \times \log_7 8 \times \log_8 9$
Using the change of base formula for logarithms, we can rewrite each term:
$\log_3 4 = \frac{\log 4}{\log 3}, \quad \log_4 5 = \frac{\log 5}{\log 4}, \quad \log_5 6 = \frac{\log 6}{\log 5}, \dots$
The product simplifies as all the intermediate logarithms cancel out, leaving:
$abcdef = \frac{\log 9}{\log 3} = 2$
Thus, the correct answer is Option (2).