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

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2

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14

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

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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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\(n=\left\lfloor log_{ 10 }(x) \right\rfloor +1\)

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

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

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