Find the first digit of $2^{100}$ without using logarithms or Newton's binomial.

Question

Find the first digit of $2^{100}$ without using logarithms or Newton's binomial.

I already have the answer, it's $1$ and the number of digits is $31$, but the proof uses logarithms and this question is for inferior grades - they didn't learn about this kind of stuff.

I think the idea is to prove $10^{30} \lt 2^{100} \lt 2 \cdot 10 ^ {30}$ , but I cannot do it without logarithms.

Any hint is appreciated.

Answer 1

We start by remarking that $2^{10}=1024=1.024\times 10^3$. It follows that $$2^{100}=(1.024)^{10}\times 10^{30}$$

Clearly $1<(1.024)^{10}$. If we can prove that $(1.024)^{10}<2$ we'll know that the first digit is $1$.

We remark that $$1.024<1.03 \implies (1.024)^2=1.048576<1.05$$

This in turn implies that $$(1.024)^4<(1.05)^2=1.1025<1.1$$

Continue: $$(1.024)^8<(1.11)^2=1.2321<1.25$$

And finally $$(1.024)^{10}=(1.024)^8\times (1.024)^2<1.25\times 1.05=1.3125<2$$ And we are done.

Answer 2

$2^{100} = (1024)^{10}\\ 10^{30}<1000^{10} <1024^{10}=2^{10}$ That establishes the lower bound.

Now we need to show that

$2^{100} < 2\times 10^{30}\\ 2^{99}< 10^{30}\\ 2^{33}<10^{10}$

$(2048)^3< (2050)^3\\ 2050\cdot 2050\cdot 2050 = (4202500)(2050) = 9\times 10^9 < 10^{10}$

$10^{30}< 2^{100} <2\times 10^{30}$

Answer 3

We need to show these "inferior grade" students that $10^{30}<2^{100}<2\cdot 10^{30}$. We compute $2^{10}=1024>10^3$. Raising both sides to the $10$th power gives the first inequality.

For the second, we note that $2^{10}<1.03 \times 10^3$. Square both sides:

$$2^{20}<1.0609 \times 10^6 < 1.1\times 10^6.$$

Square again:

$$2^{40}<1.21 \times 10^{12} <1.3 \times 10^{12}.$$

Square again:

$$2^{80} < 1.69 \times 10^{24} < 1.7 \times 10^{24}.$$

Then compute:

$$2^{100} =2^{80}2^{20} < 1.7\times 10^{24}\times 1.1\times 10^6 =1.87\times 10^{30}<2\times 10^{30}.$$