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# Power

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 · 2 years, 3 months ago

2 · 1 year, 8 months ago

14 · 1 year, 8 months ago

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

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

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

The answer is $$\log {2^{2^{22}}}$$ · 2 years, 3 months ago

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

I forgot to put it in Round function · 2 years, 3 months ago

You forgot to add $$1$$. Floor function is understood. · 2 years, 3 months ago

I meant Round off function · 2 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:

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

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

Is it always true ? · 2 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. · 2 years, 3 months ago