If $x^{100}$ is 31 digit number Then $x^{1000}$ contains how many digits.
Our Approach:
$10^1$ has $2$ digits = $10$
$10^2$ has $3$ digits = $100$
$10^3$ has $4$ digits = $1000$
$10^{30}$ has $31$ digits
therefore, $10^{30}=x^{100}$
$30 = 100\log x$
multiplying both sides by $10$
$30 \cdot 10 = 10 \cdot 100\log x$
$300 = 1000\log x = 300+1=301$
so $301$ is correct answer
Q Any other approach to solve this problem and Also correct me if i am wrong.
Best Answer
$x^{100}$ is 31 digit => $$10^{30}\leq x^{100}< 10^{31} \Rightarrow 10^{300}\leq x^{1000}< 10^{310}.$$ This means that $x^{1000}$ has from 301 t0 310 digits.