2025 AMC 12A 第 7 题

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7.

在某个外星世界中,生物的最大奔跑速度 vv 取决于它的脚趾数 nn 和眼睛数 mm。这种关系可以表示为 v=knambv = k n^a m^b 厘米每小时,其中 kkaa, 和 bb 是整数常数。在所有生物都有 55 个脚趾的群体中,logv=4+2logm\log v = 4 + 2\log m; 在所有生物都有 2525 只眼睛的群体中,logv=4+4logn\log v = 4 + 4\log n, 其中对数均以 1010 为底。k+a+bk + a + b 是多少?

In a certain alien world, the maximum running speed vv of an organism is dependent on its number of toes nn and number of eyes m.m. The relationship can be expressed as v=knambv = k n^a m^b centimeters per hour, where k,k, a,a, and bb are integer constants. In a population where all organisms have 55 toes, logv=4+2logm;\log v = 4 + 2\log m; and in a population where all organisms have 2525 eyes, logv=4+4logn,\log v = 4 + 4\log n, where the logarithms are base 10.10. What is k+a+b?k + a + b?

2020

2121

2222

2323

2424

答案:C
知识点:对数方程组
难度评级:1380
解答:

取对数,logv=logk+alogn+blogm\log v = \log k + a\log n + b\log m

n=5n = 5 时,式子变为 logv=(logk+alog5)\log v = (\log k + a\log 5) +blogm+ b\log m4+2logm4 + 2\log m, 对应,所以 b=2b = 2,且 logk+alog5=4\log k + a\log 5 = 4

m=25m = 25 时,式子变为 logv=(logk+blog25)\log v = (\log k + b\log 25) +alogn+ a\log n4+4logn4 + 4\log n, 对应,所以 a=4a = 4,且 logk+2log25=4\log k + 2\log 25 = 4

因此 logk=4log625\log k = 4 - \log 625 =log10000625= \log\dfrac{10000}{625} =log16= \log 16, 所以 k=16k = 16。于是 k+a+b=16+4+2=22k + a + b = 16 + 4 + 2 = 22

因此,正确答案是 C

Taking logarithms, logv=logk+alogn+blogm.\log v = \log k + a\log n + b\log m.

With n=5,n = 5, this reads logv=(logk+alog5)\log v = (\log k + a\log 5) +blogm,+ b\log m, matching 4+2logm,4 + 2\log m, so b=2b = 2 and logk+alog5=4.\log k + a\log 5 = 4.

With m=25,m = 25, it reads logv=(logk+blog25)\log v = (\log k + b\log 25) +alogn,+ a\log n, matching 4+4logn,4 + 4\log n, so a=4a = 4 and logk+2log25=4.\log k + 2\log 25 = 4.

Then logk=4log625\log k = 4 - \log 625 =log10000625= \log\dfrac{10000}{625} =log16,= \log 16, so k=16.k = 16. Hence k+a+b=16+4+2=22.k + a + b = 16 + 4 + 2 = 22.

Thus, the correct answer is C.

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