April 12th, 2018, 08:13 AM  #11 
Senior Member Joined: May 2016 From: USA Posts: 989 Thanks: 406  Yes $1 \le a < 10,\ a \in \mathbb R,\ n,\ x \in \mathbb Z,\ n \ge 0, \text { and } x = a * 10^n \implies$ $\text { there are exactly } n + 1 \text { digits in } x.$ $2.03 * 10^{90} < 2^{300} < 2.04 * 10^{90} \implies$ $(2.03 * 10^{90})^4 < (2^{300})^4 < (2.04 * 10^{90})^4 \implies$ $2.03^4 * 10^{360} < 2^{1200} < 2.04^4 * 10^{360}) \implies$ $ 16.98 * 10^{360} < 2^{1200} < 17.32 * 10^{360}) \implies$ $2^{1200} \approx 17 * 10^{360} = 1.7 * 10^{361} \implies$ $x \text { has exactly } 361 + 1 = 362 \text { digits.}$ 

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