@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.
–
Raghav Vaidyanathan
·
2 years, 1 month ago

## Comments

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

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2 – Joe Doe · 1 year, 6 months ago

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14 – Joe Doe · 1 year, 6 months ago

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Is there any way of calculation without using logarithms? – Swapnil Das · 2 years, 1 month ago

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– David Holcer · 2 years, 1 month ago

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

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

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The answer is \( \log {2^{2^{22}}}\) – Rajdeep Dhingra · 2 years, 1 month ago

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– Raghav Vaidyanathan · 2 years, 1 month ago

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

– Rajdeep Dhingra · 2 years, 1 month ago

I forgot to put it in Round functionLog in to reply

– Raghav Vaidyanathan · 2 years, 1 month ago

You forgot to add \(1\). Floor function is understood.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

I meant Round off functionLog in to reply

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

\(n\ne \left\lfloor log_{ 10 }(x) \right\rfloor\) – Raghav Vaidyanathan · 2 years, 1 month ago

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– Rajdeep Dhingra · 2 years, 1 month ago

Is it always true ?Log in to reply

In general: \(10^k\) has \(\left\lfloor k \right\rfloor+1\) digits. – Raghav Vaidyanathan · 2 years, 1 month ago

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