1951 AMC 12 第 35 题

Question

若

a^x=c^q=b

，且

c^y=a^z=d

，则：

If

a^x=c^q=b

and

c^y=a^z=d,

then:

xy=qz

\dfrac{x}{y}=\dfrac{q}{z}

x+y=q+z

x-y=q-z

x^y=q^z

对两组连等式分别取对数

Take logarithms of both chains of equal powers

由

x\log a=q\log c

与

y\log c=z\log a

消去对数因子

From

x\log a=q\log c

and

y\log c=z\log a

, eliminate the logarithms

取对数，得到

\begin{gathered} x\log a=q\log c,\\ y\log c=z\log a \end{gathered}\text{。}

按照本题历史上对非退化底数的约定，将两式相乘并约去非零的对数因子，得到

xy=qz

。

因此，题目预期的正确答案是 A。

Taking logarithms gives

\begin{gathered} x\log a=q\log c,\\ y\log c=z\log a. \end{gathered}

Multiplying these equations and cancelling the nonzero logarithmic factors under the historical nondegenerate-base convention yields

xy=qz.

Thus, the intended correct answer is A.

其他年份的第 35 题