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I've been trying to find a solution to this problem for a while but I just can't seem to find the connection between the numbers and I really need help. I apologize if a problem like this one has appeared already.

If I know that log102 = 0,3010 how do I find out how many digits there are in the number 529?

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The number of digits in a number $n$ is $1+\lfloor\log_{10}n\rfloor$, so it suffices to compute $\log_{10}(5^{29})$. But

$$\log_{10}(5^{29})=29\log_{10}5=29\log_{10}(10/2)=29(1-\log_{10}2)$$.

Can you take it from there?

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$$5^{29}=10^{29\log_{10}5}=10^{29(1-\log_{10}2)}=10^{29(1-0.3010)}$$ So, your number have $\lfloor{29 (1 - 0.3010)}\rfloor+1=21$ digits.

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  • $\begingroup$ @RossMillikan. Yes, you are right. I edited question. $\endgroup$ – user164524 May 28 '15 at 16:22

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