@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
·
1 year, 9 months ago

## Comments

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TopNewest\(\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1\)= 1262612 – Raghav Vaidyanathan · 1 year, 9 months ago

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2 – Joe Doe · 1 year, 1 month ago

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14 – Joe Doe · 1 year, 1 month ago

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Is there any way of calculation without using logarithms? – Swapnil Das · 1 year, 9 months ago

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– David Holcer · 1 year, 9 months 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 · 1 year, 9 months ago

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The answer is \( \log {2^{2^{22}}}\) – Rajdeep Dhingra · 1 year, 9 months ago

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– Raghav Vaidyanathan · 1 year, 9 months ago

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

– Rajdeep Dhingra · 1 year, 9 months ago

I forgot to put it in Round functionLog in to reply

– Raghav Vaidyanathan · 1 year, 9 months ago

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

– Rajdeep Dhingra · 1 year, 9 months 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 · 1 year, 9 months ago

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– Rajdeep Dhingra · 1 year, 9 months ago

Is it always true ?Log in to reply

In general: \(10^k\) has \(\left\lfloor k \right\rfloor+1\) digits. – Raghav Vaidyanathan · 1 year, 9 months ago

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