Large power of 2

We can determine first digits of large numbers by finding their log values to base 10. Because powers of 10 will have whole number log values, numbers with a first digit of 1 will have log values slightly above a whole number, whereas numbers with a high first digit will have log values just below a whole number.

We have $\log_{10} 2^{10} \approx 3.0103$ , so $\log_{10} 2^{10N} = N\log_{10} 2^{10} \approx 3.0103N$ . What we're looking for is a value of $N$ that will push this as close to a whole number as possible without going over (like The Price Is Right!), so that the next $N$ will push it just past a whole. Since the first two decimal digits are $01$ , we consider something close to $N=99$ , but to account for the digits $03$ that follow, we start slightly lower. We find that

$\begin{aligned} \bullet \ N&=96 \implies 96\log_{10} 2^{10} \approx 288.9888 \\ \bullet \ N&=97 \implies 97\log_{10} 2^{10} \approx 291.9991 \\ \bullet \ N&=98 \implies 98\log_{10} 2^{10} \approx 295.0094 \end{aligned}$

Verifying $2^{970} \approx 9.9792 \times 10^{291}$ and $2^{980} \approx 1.0219 \times 10^{295}$ so the value of $N$ we're looking for is $\boxed{98}$

My method: $10Nlog_{10}{2}$ will help write the number in base $10$ . Let $\{10Nlog_{10}{2}\}$ be the mantissa (the decimal part.) Then $10^{\{10Nlog_{10}{2}\}}$ will have the same first digit as $2^{10N}$ . Make a simple table. $N=30$ is the first that doesn't begin with $1$ . Continue to $N=97$ , which ends in 9 and right after that $\boxed{N=98}$ begins with 1 again.