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robkelly2242 · Feb 26, 2008, 04:06 PM

If x = log 2 and y = log 3, all logs to base 10 and to 3 significant figures, what are the logs of each number 1 through 10 in terms of x and y?

ebaines · Feb 27, 2008, 11:02 AM

This is straight-forward for those numbers that are pure multiples of 2 and/or 3. For example: log(6) = log(2*3) = log(2) + log(3) = x + y; and log(8) = log(2^3) = 3 log(2) = 3x. But you can't express log(5), log(7), or log(10) in terms of log(2) and log(3) only. Is there something else in this problem you aren't telling us? Or am I misreading it?

robkelly2242 · Feb 27, 2008, 01:47 PM

Ebaines, log1 = 0, by definition. Log10 = 1, again by definition. Since 5 = 10/2, log 5 = 1 minus x(log 10 minus log 2). If you want to be a purist, and limit yourself only to x and y, substitute x-x for 0, and x/x for 1. Log 7? Ah, there's the rub! Hint: 7 is a prime #. You have already dealt with the needed concept in expressing logs 4, 8 and 9. Rob

ebaines · Feb 27, 2008, 02:53 PM

Ah.. I completely forgot that these are log base 10 - doh! Sometimes the simplest things are the hardest to see. But I'm still flumoxed by log(7). 7 is indeed a prime number, hence can't be expressed in terms of multiples of 2, 3, 5 etc. So I'm stuck - what am I missing?

robkelly2242 · Feb 27, 2008, 03:10 PM

The other simple thing you didn't take into account is that these are logs to only three significant figures. So log7 is (2 + y + 3x)/4. Rob

ebaines · Feb 27, 2008, 03:26 PM

Cute! No way was I going to see that 7^4 is so close to a number that is a factor of only 2, 3, and 5. I had thought about the fact that 7 is about halfway between 10/(cube root3) and 10/(square root 2). If you average those you get: 1-y/6 -x/4, which is close enough.