If ` log_(10) 2 = 0.3010 and log_(10) 3 = 0.477`, then find the number of digits in the following numbers:
(a)` 3^(40)" "(b) 2^(32) xx 5^(25)" (c) 24^(24)`

Answer

(a) 20 (b) 28 (c) 34

Explanation

(a) ` N = 3^(40)`
`:. Log_(10)N = 40 log_(10) 3 = 40 xx 0.477 = 19.08`
So, number of digits in N is 20.
(b) `N= 2^(32) xx 5^(25) = 2^(7) (2 xx 5)^(25) = 2^(7) xx 10^(25)`
` :. Log_(10) N = 25 + 7 log_(10) 2`
` = 25 + 7 xx 0.3010`
` = 25+ 2.107`
` = 27.107`
So, number of digits in N is a 28.
(c) `log_(10)24^(24) = 24 (log_(10)(8xx3))`
` = 24 [3 log_(10)2+log_(10)3]`
` 24[3 xx 0.3010+ 0 .477]`
` = 24(1.38)`
` = 33.12`
So, number of digits in N is 34.