I found a couple of questions where, for example, they ask you to calculate the number of digits in $18^{200}$ and only the value of $\log 18$ is given. Can anyone tell me a way?
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Assuming the logarithm is in base 10, if you take the log of any number in base 10 and round it up, you will get the number of digits in that number. – ignoramus Aug 05 '15 at 03:26
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Log of a number returns approximately the number of digits. Log 1=0, log 10=1, log 100=2 etc. Log(18^200) = 200.log(18). – Dr Xorile Aug 05 '15 at 03:26
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1The number of decimal digits in $N$ would be $\lfloor \log_{10} N \rfloor+1$. – Macavity Aug 05 '15 at 03:26
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1Can you guys tell me how we get this result? – Soham Ghosh Aug 05 '15 at 03:35
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Check http://math.stackexchange.com/questions/231742/proof-how-many-digits-does-a-number-have-lfloor-log-10-n-rfloor-1 – Macavity Aug 05 '15 at 03:40
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Thanks! That was exactly what I was looking for. :) – Soham Ghosh Aug 05 '15 at 03:44
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It is not too tough to find the # of digits in $200\times\ log_{10}(18)$ = 251.05...
You might recall that the integer portion of the logarithm is called the characteristic, and the decimal part, the mantissa
You would also recall that: log(10) = 1, and the logarithm remains less than 2 as long as N<100, log(100) = 2, and so on.
The simple rule that emerges is: Add 1 to the characteristic to get the # of digits in the number.
Or, in briefer mathematical notation, using the floor function, D = $\lfloor \log_{10} N \rfloor+1$
true blue anil
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1You should know the basic rules of logs ! log($a\cdot b) = log(a) + log(b), log(a/b) = log(a) - log(b), and log$(a^b)$ = b*log(a) – true blue anil Aug 05 '15 at 03:41
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2Exactly how does the number of digits in $200\times \log 18$ answer the problem? That's wrong. – Macavity Aug 05 '15 at 03:45
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