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Find the number of digits in \({ 2 }^{ 2^{ 22 } }\)

Note by Swapnil Das
2 years, 3 months ago

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\(\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1\)= 1262612 Raghav Vaidyanathan · 2 years, 3 months ago

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2 Joe Doe · 1 year, 8 months ago

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14 Joe Doe · 1 year, 8 months ago

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Is there any way of calculation without using logarithms? Swapnil Das · 2 years, 3 months ago

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@Swapnil Das yes, we could try writing it as 10 to a certain power(changing base/power.) David Holcer · 2 years, 3 months ago

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22log2+1 = 7 (log2 =.3010) Nithin Nithu · 2 years, 3 months ago

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The answer is \( \log {2^{2^{22}}}\) Rajdeep Dhingra · 2 years, 3 months ago

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@Rajdeep Dhingra No, it is not, it is \(\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1\) Raghav Vaidyanathan · 2 years, 3 months ago

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@Raghav Vaidyanathan I forgot to put it in Round function Rajdeep Dhingra · 2 years, 3 months ago

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@Rajdeep Dhingra You forgot to add \(1\). Floor function is understood. Raghav Vaidyanathan · 2 years, 3 months ago

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@Raghav Vaidyanathan I meant Round off function Rajdeep Dhingra · 2 years, 3 months ago

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@Rajdeep Dhingra I get the round off function bit. The thing that I meant is that the number of digits \(n\) of a natural number \(x\) is given by:

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

\(n\ne \left\lfloor log_{ 10 }(x) \right\rfloor\) Raghav Vaidyanathan · 2 years, 3 months ago

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@Raghav Vaidyanathan Is it always true ? Rajdeep Dhingra · 2 years, 3 months ago

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@Rajdeep Dhingra 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. Raghav Vaidyanathan · 2 years, 3 months ago

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