请教OG-11的一道数学题

quanquanwins

请教一道数学题，看都没看明白。如能解释意思和如何解答，不胜感激。

On a scale that measures the intensity of a certain phenomenon, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?

goodfishs

某种现象密度的比例尺，n+1代表密度为10*n，8=7+1代表7*10=70，3=2+1代表2*10=20，所以倍数关系为70/20=3.5倍

不知是否正确？

freedownload

QUOTE:

请教一道数学题，看都没看明白。如能解释意思和如何解答，不胜感激。

On a scale that measures the intensity of a certain phenomenon, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?

10^5?

quanquanwins

答案是10的5次幂也。请教楼主如何得出？

quejingui77

A reading of 8 is 10 times that of 7, which is 10 times that of 6, which is 10 times that of 5, which is 10 times that of 4, which 10 times that of 3. 一共5个10 times，前后递进关系，所以是10^5。

又或者可以这样想，每一个reading的实际值是C*10^n（C是任意一个非零常数），这样n+1和n的读数就像题目说的那样相差10倍。于是8的实际值是C*10^8，3的实际值是C*10^3，所以两者相差的倍数就是(C*10^8)/(C*10^3)=10^5

XIAOSHAONIAN

DING !![em01]

quanquanwins

thanks  a  lot !!!!!!!!!!!!!!!!!!!!