2005 AMC 10B Problem 17

Question

Suppose that

4^a = 5,

5^b = 6,

6^c = 7,

and

7^d = 8.

What is

a \cdot b \cdot c \cdot d?

1

\dfrac32

2

\dfrac52

3

Raise successive equations to powers so the exponents multiply

4^{abcd}

telescopes down to

8;

write

8

as a power of

4

Chaining the equations,

\begin{aligned} 4^{abcd} &= \left(\left(\left(4^a\right)^b\right)^c\right)^d \\ &= \left(\left(5^b\right)^c\right)^d \\ &= \left(6^c\right)^d = 7^d = 8. \end{aligned}

Since

8 = 4^{\frac{3}{2}},

we conclude

a \cdot b \cdot c \cdot d = \dfrac32.

Thus, B is the correct answer.

Problem 17 in other years