AMC12 2010 A

AMC12 2010 A · Q11

AMC12 2010 A · Q11. It mainly tests Exponents & radicals, Logarithms (rare).

The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \log_b 7^7$. What is $b$?

方程 $7^{x+7} = 8^x$ 的解可以表示为 $x = \log_b 7^7$ 的形式。$b$ 的值是多少？

(A) $\frac{7}{15}$ $\frac{7}{15}$

(B) $\frac{7}{8}$ $\frac{7}{8}$

(D) $\frac{15}{8}$ $\frac{15}{8}$

(E) $\frac{15}{7}$ $\frac{15}{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

This problem is quickly solved with knowledge of the laws of exponents and logarithms. \begin{align*} 7^{x+7} &= 8^x \\ 7^x*7^7 &= 8^x \\ \left(\frac{8}{7}\right)^x &= 7^7 \\ x &= \log_{8/7}7^7 \end{align*} Since we are looking for the base of the logarithm, our answer is $\boxed{\textbf{(C)}\ \frac{8}{7}}$.

本题可利用指数与对数的运算律快速求解。 \begin{align*} 7^{x+7} &= 8^x \\ 7^x*7^7 &= 8^x \\ \left(\frac{8}{7}\right)^x &= 7^7 \\ x &= \log_{8/7}7^7 \end{align*} 因为我们要找的是对数的底数，所以答案是 $\boxed{\textbf{(C)}\ \frac{8}{7}}$。

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