U may assume the no. Of digits of a number be x. Let the number be a^b. So
10^(x-1) = a^b
X-1 = blog(a) [log with base 10]
X = blog(a) +1
For convenience, X = [blog(a)] +1
So 2^22 has [22log(2)] +1 digits.similarly following can be calculat ed

@Aditya Narayan Sharma
–
How to apply here .here it will become 2^22 log(2) then we have to apply log again then do antilog.after doing this I got answer 447 .is it right??????

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest1262612

Log in to reply

Actually I got the answer but according to my book answer key I was wrong so....Thanks

Log in to reply

@Dev Sharma DevBut how will you find the number of digits without computing

\(\large2^{22}\)if it is asked in exam?Log in to reply

U may assume the no. Of digits of a number be x. Let the number be a^b. So 10^(x-1) = a^b X-1 = blog(a) [log with base 10] X = blog(a) +1 For convenience, X = [blog(a)] +1 So 2^22 has [22log(2)] +1 digits.similarly following can be calculat ed

Log in to reply

Log in to reply

@Calvin Lin Please help me with this!

Log in to reply