The number of digits in 20^{301} (given, log | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $y = 20^{301}$.To find the number of digits in $y$, we calculate $\lfloor \log_{10} y \rfloor + 1$.$\log_{10} y = \log_{10} (20^{301}) = 301 \

Vedclass · Accepted Answer

$392$

Vedclass · Answer

(C) Let $y = 20^{301}$.To find the number of digits in $y$, we calculate $\lfloor \log_{10} y floor + 1$.$\log_{10} y = \log_{10} (20^{301}) = 301 	imes \log_{10} (2 	imes 10)$.Using the property $\log(ab) = \log a + \log b$, we get:$\log_{10} y = 301 	imes (\log_{10} 2 + \log_{10} 10) = 301 	imes (0.3010 + 1) = 301 	imes 1.3010$.$301 	imes 1.3010 = 391.601$.The number of digits is $\lfloor 391.601 floor + 1 = 391 + 1 = 392$.

The number of digits in $20^{301}$ (given, $\log _{10} 2=0.3010$) is:

Similar Questions

If $a < 1$ and $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x=$

If $x = \log_3 5$ and $y = \log_{17} 25$,then which of the following is correct?

If ${\log _5}a \cdot {\log _a}x = 2$,then the value of $x$ is:

If $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$,then the relationship between $a$ and $b$ is:

If $a, b, c$ are distinct positive numbers,each different from $1$,such that $[(\log _b a)(\log _c a) - (\log _a a)] + [(\log _a b)(\log _c b) - (\log _b b)] + [(\log _a c)(\log _b c) - (\log _c c)] = 0$,then $abc =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module