How do I know how many digits I am off by when calculating $200^{2010}$ or $125^{100}$?

Question

I want to solve this problem without the use of logarithms. If so this is how I solve it for the first question:

$200^{2010} = (100 * 2) ^{2010} = (10^{2} * 2)^{2010} = (10^{2})^{2010}*2^{2010} = 10^{4020}*(2^{10})^{201} \approx 10^{4020}*(10^{3})^{201} = 10^{4623}$

So there would be $4624$ digits since my approximation rounds down the value of $2^{10}$.

But the answer is $4626$, how can I account for this with my approximation?

Similarly $125^{100} = \left(\frac{1000}{8}\right)^{100} = \left(\frac{10^{3}}{2^{3}}\right)^{100}$ Using the same approximation of $2^{10} \approx 10^3$ I would get

$$\frac{10^{300}}{\left(10^{3}\right)^{30}} = \frac{10^{300}}{10^{90}} = 10^{210}$$

But I know my approximation leaves me with one digit off so the answer is $211$ digits.

My question is, why is my approximation wrong in the first question and not wrong in the second question? And how can I account for these errors without the use of logarithms?

Answer 1

Since $2^{10} = 10^3 \cdot \frac{2^{10}}{10^3}$, $({2^{10}})^{201} = ({10^3})^{201} \cdot (\frac{2^{10}}{10^3})^{201}$. The error $(\frac{2^{10}}{10^3})^{201}$, or $1.024^{201}$ is approximately $1.025^{200}$, or $(1+\frac{1}{40})^{200}$.

Using $(1+\frac{1}{40})^{40} \approx e$, then $(1+\frac{1}{40})^{200}$ is just $\big(1+\frac{1}{40})^{40}\big)^5$ or $e^5$. Since $2.7 < e < 3$, by direct computation, $100 < (\frac{27}{10})^5$, and $3^5 < 1000$, so $ 100 < (\frac{5}{2})^5 < 1000$. Therefore, the actual answer will be $2$ digits longer than your approximation.

Answer 2

For better estimation, use $\log_{10}2 \approx 0.30103$ as pointed out by @Toby Mak. You do not need a calculator, even for fairly large numbers. For example: How many digits for $2^{10000}$?: $$\log_{10}2^{10000} = 10000\left(log_{10}2\right) = 10000\times0.30103\approx30104 $$ In your problem, it was $4020+2010\times0.30103\approx4020+606=4626$, the correct answer.

Similarly, this would work for your second problem as well. But keep in mind that in $\log$ calculations, you may need to subtract a digit during divisions rather than adding one we did during multiplication: $$125^{100}=\left(\frac{1000}{8}\right)^{100}=\left(\frac{10^{3}}{2^{3}}\right)^{100}=\frac{10^{300}}{2^{300}}$$ $$\log_{10}2^{300} = 300\left(\log_{10}2\right) = 300\times0.30103\approx90.309 $$ it wasTherefore, number of digits should be: $300-(90-1)=211$, the correct answer.