# Power

Find the number of digits in $${ 2 }^{ 2^{ 22 } }$$

Note by Swapnil Das
3 years, 3 months ago

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

- 3 years, 3 months ago

2

- 2 years, 8 months ago

14

- 2 years, 8 months ago

Is there any way of calculation without using logarithms?

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

I forgot to put it in Round function

- 3 years, 3 months ago

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

- 3 years, 3 months ago

I meant Round off function

- 3 years, 3 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:

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

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

- 3 years, 3 months ago

Is it always true ?

- 3 years, 3 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.

- 3 years, 3 months ago