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2005 AMC 10B Problem 17

Problem 17 of 25IntermediateAlgebra

Suppose that 4a=5,4^a = 5, 5b=6,5^b = 6, 6c=7,6^c = 7, and 7d=8.7^d = 8. What is abcd?a \cdot b \cdot c \cdot d?

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Solution

Chaining the equations, 4abcd=(((4a)b)c)d=((5b)c)d=(6c)d=7d=8. \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=432,8 = 4^{\frac{3}{2}}, we conclude abcd=32.a \cdot b \cdot c \cdot d = \dfrac32. Thus, B is the correct answer.

More practice

Concepts: exponent · telescoping

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.