2005 AMC 10B Problem 17

Each equation expresses a number as a power of the previous one, so nest them. Start from $8$:
$$8 = 7^d.$$
Replace $7$ by $6^c$: $8 = (6^c)^d = 6^{cd}$.

Replace $6$ by $5^b$: $8 = (5^b)^{cd} = 5^{bcd}$.

Replace $5$ by $4^a$: $8 = (4^a)^{bcd} = 4^{abcd}$.

Now both sides are powers of $2$: $2^3 = (2^2)^{abcd} = 2^{2abcd}$. Equal bases force equal exponents, so $2abcd = 3$ and
$$abcd = \frac{3}{2}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{3}{2}}$.

A chain of the form $x_1^{e_1} = x_2$, $x_2^{e_2} = x_3$, ... collapses by repeated substitution into $x_1^{e_1 e_2 \cdots} = x_{\text{last}}$; the product of the exponents is the natural quantity, which is why the problem asks for it. In log language, $abcd = \log_4 8$, and the intermediate bases cancel like a telescoping product.

Write $a = \log_4 5$, $b = \log_5 6$, $c = \log_6 7$, $d = \log_7 8$. By the change-of-base formula the product is $\frac{\ln 5}{\ln 4}\cdot\frac{\ln 6}{\ln 5}\cdot\frac{\ln 7}{\ln 6}\cdot\frac{\ln 8}{\ln 7} = \frac{\ln 8}{\ln 4} = \frac{3}{2}$.

• Trying to compute a, b, c, d individually, or ending with 4^{abcd} = 8 and answering 2 by comparing 8 to 4 wrongly.
• Adding exponents instead of multiplying when simplifying $(6^c)^d$.