How many digits are in $125^{100}$?

Question

What I can think of thus far is that $125^{100} = (\frac{1000}{8})^{100} = \frac{1000^{100}}{2^{300}}$

I know that $2^{10} = 1024$ so $\frac{1000^{100}}{1024^{30}}$.

That's all I can figure out this far.

I was thinking to divide the numerator and denominator of $\frac{1000^{100}}{1024^{30}}$ by $1000^{30}$ and I think that would give me $\frac{1000^{70}}{1.024^{30}}$ but I'm not even sure if this is correct.

Can someone please help me solve this?

Edit: How can I solve this without the use of logarithms?

Answer 1

$2^{10}\approx 10^3$, so approximately, $\frac {1000^{100}}{10^{90}} =\frac {100^{100}\cdot 10^{100}}{10^{90}}=100^{100}\cdot 10^{10}=10^{210}$... So about $211$.

Answer 2

Logarithm base 10 Scale will better answer your question.
Range                        Digits
1-log(10)                        1
log(10)-log(10^2)        2
log(10^2)-log(10^3)   3
log(10^3)-log(10^4)   4
log(10^n-1)-log(10^n)  n
I think you get the Idea.

Now for your number $$125^{100} = 5^{300}$$

$$\log_{10}{5^{300}} = 300 * \log_{10}{5} = 209.691$$

That means it is between 209 and 210. From the above table pattern, you can confirm there will be 210 digits.

Answer 3

Calculate

$\log_{10}125^{100}$

$= 100\cdot \log_{10} (1000/8)$

$= 100\cdot(3-3\log_{10}2)$

$= 100\cdot(3-0.9030)$

$= 100\cdot(2.0970)$

$= 209.70$

Therefore number of digits $= [209.70]+1 = 210.$

Answer 4

Any solution implicitly computes a logarithm ($\log_{10} 5$ is approximately the answer to this question divided by $300$), but one can correct Chris Custer's answer for the extra digit by recalling the tangentially logarithmic "rule of $72$": An interest rate of $2.4$ percent will double the principle after $72/2.4 = 30$ periods. So $(2^{10})^{30}$ will approximately equal $2 \times 10^{90}$, rather than just $10^{90}$. That cuts the number of digits from $211$ to $210$.

Answer 5

For any natural number n the number of digits is $ 1+[log(n)]$ where $ [log(n)]$ stands for the integer part of $log(n)$

For $n=125^{100}$, we get $$log(n)= 100 log(125)=209.6910013...$$

Thus the number of digits in $125^{100}$ is $210.$

Answer 6

For all $x$ we have $$125^{100} < 125^{x}128^{100-x}=5^{3x}2^{700-7x}$$

Making $700-7x=3x$ we get $x=70$ and hence $$125^{100} < 5^{210}2^{210}=10^{210}$$ [I could had compared these two numbers directly, but using the $x$ above shows how I figured out that 210 is the key].

Since the approximation above is very close, intuition tells us that $125^{100}$ should have 210 digits. But intuiton is not enough, we need to prove it. Since we showed $125^{100} <10^{210}$, to conclude that $125^{100}$ has 210 digits we need to show $$125^{100} \geq 10^{209}$

Lets prove it $$125^{100} \geq 10^{209} \Leftrightarrow \\ 5^{300} \geq 2^{209} 5^{209} \Leftrightarrow \\ 5^{91} \geq 2^{209} \Leftrightarrow \\ 5^{13} \geq 2^{29.86}$$

The last inequality holds since $5^{13}>2^{30}$, but I don't see any simple way of getting this without a calculator.

Answer 7

As you worked out, $125^{100}=\dfrac{10^{210}}{1.024^{30}}$, so it is enough to show $1<1.024^{30}<10$ to conclude there are $210$ digits. But the left inequality is obvious and as $1+x\leqslant e^x$, we get $1.024^{30}\leqslant e^{0.024\times 30}=e^{0.72}<e<10,\;$ so the right inequality holds true as well.