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@Rajdeep Dhingra
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Yes, this is due to the fact that $log_{10}(x)$ gives us the number of zeroes in the largest power of $10$ which is less than $x$. For example, if $x=12345$, $log_{10}(x)=4$(approx.) as $10000$ is the largest power of ten below $12345$. But we need to add one to this value to account for the leading number as we have now counted only the zeroes and left out the one in $10000$. Therefore, the right answer is $4+1=5$.

In general: $10^k$ has $\left\lfloor k \right\rfloor+1$ digits.

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## Comments

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TopNewest$\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1$= 1262612

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The answer is $\log {2^{2^{22}}}$

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No, it is not, it is $\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1$

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I forgot to put it in Round function

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$1$. Floor function is understood.

You forgot to addLog in to reply

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$n$ of a natural number $x$ is given by:

I get the round off function bit. The thing that I meant is that the number of digits$n=\left\lfloor log_{ 10 }(x) \right\rfloor +1$

$n\ne \left\lfloor log_{ 10 }(x) \right\rfloor$

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$log_{10}(x)$ gives us the number of zeroes in the largest power of $10$ which is less than $x$. For example, if $x=12345$, $log_{10}(x)=4$(approx.) as $10000$ is the largest power of ten below $12345$. But we need to add one to this value to account for the leading number as we have now counted only the zeroes and left out the one in $10000$. Therefore, the right answer is $4+1=5$.

Yes, this is due to the fact thatIn general: $10^k$ has $\left\lfloor k \right\rfloor+1$ digits.

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22log2+1 = 7 (log2 =.3010)

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Is there any way of calculation without using logarithms?

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yes, we could try writing it as 10 to a certain power(changing base/power.)

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14

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2

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