# Power

Find the number of digits in ${ 2 }^{ 2^{ 22 } }$ Note by Swapnil Das
4 years, 9 months ago

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

- 4 years, 9 months ago

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

- 4 years, 9 months ago

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

- 4 years, 9 months ago

I forgot to put it in Round function

- 4 years, 9 months ago

You forgot to add $1$. Floor function is understood.

- 4 years, 9 months ago

I meant Round off function

- 4 years, 9 months ago

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:

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

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

- 4 years, 9 months ago

Is it always true ?

- 4 years, 9 months ago

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.

- 4 years, 9 months ago

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

- 4 years, 9 months ago

Is there any way of calculation without using logarithms?

- 4 years, 9 months ago

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

- 4 years, 9 months ago

14

- 4 years, 1 month ago

2

- 4 years, 1 month ago