What is the smallest power of $2$ with at least $1,000,000,000$ digits in base-$10$?

Question

$$2^x > 10^{1000000000}$$

Smallest integer value of $x$.

I can't seem to find a calculator that can calculate those high numbers and I'm not sure of any way to figure this out other than guess and check.

Answer 1

This is easy to calculate taking logarithms. With logarithms in base $b$ you know the number of digits of the input ($2^{x}$) in that base. In this case base $10$.

$\log_{10}(2^{x}) > 1,000,000,000$

$x\log_{10}(2) > 1,000,000,000$

$x > \frac{1,000,000,000}{\log_{10}(2)}$

x > 3321928094.887362348

Answer 2

Since logarithm is increasing we have $$ x = \log_2 2^x \geq \log_2 10^{1,000,000,000} = 1,000,000,000 \log_2 10 $$ This should be doable with a calculator.

Answer 3

Take logarithms. $2^x > 10^y$ if and only if $x \log_{10}(2) > y$, i.e. $x > y/\log_{10}(2)$. In this case $$10^9/\log_{10}(2) \approx 3.3219280948873623479 \times10^9$$ so you want $x \ge 3321928095$.