AMC12 2005 B

AMC12 2005 B · Q13

AMC12 2005 B · Q13. It mainly tests Exponents & radicals.

Suppose that $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$. What is $x_1x_2...x_{124}$?

假设 $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$。$x_1x_2...x_{124}$ 等于多少？

(A) 2 2

(B) \frac{5}{2} \frac{5}{2}

(D) \frac{7}{2} \frac{7}{2}

(E) 4 4

Answer

Correct choice: (D)

正确答案：（D）

Solution

We see that we can re-write $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$ as $\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128$ by using substitution. By using the properties of exponents, we know that $4^{x_1x_2...x_{124}}=128$. $4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}$ Therefore, the answer is $\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$

我们可以用代换把 $4^{x_1}=5$, $5^{x_2}=6$, $6^{x_3}=7$, ... , $127^{x_{124}}=128$ 改写为 $\left(...\left(\left(\left(4^{x_1}\right)^{x_2}\right)^{x_3}\right)...\right)^{x_{124}}=128$。利用指数性质可知 $4^{x_1x_2...x_{124}}=128$。 $4^{x_1x_2...x_{124}}=128\\2^{2x_1x_2...x_{124}}=2^7\\2x_1x_2...x_{124}=7\\x_1x_2...x_{124}=\dfrac{7}{2}$ 因此答案为 $\boxed{\mathrm{(D)}\,\dfrac{7}{2}}$

Topics

AL.008 Exponents & radicals