how many places for int $5^{29}$

Question

how many places for int $5^{29}? \ \log_{10}{2} = 0.30$

i thought that places = digits

so, i did this

$ \log_{10}{5} = 1 - 0.3 = 0.7$

$ \log_5{10} = \frac{10}{7}$

$ 5^{10/7} = 10$

$5^{29} = 10^{203/10} = 10^{20.3}$ since, $10^2 = 100$ (2 places)

it means $10^{20.3}$ 20 or 21 places? and what $10^{20.3}$ looks like without calculator?

Answer 1

At the level of your approximations $10^{0.3}=2$, so $10^{20.3}=2\cdot 10^{20}$. $\ \ 10^{20}$ is a $1$ with $20$ zeros after it, so you would say $5^{29}$ is a $2$ with $20$ zeros after it. That means $21$ digits total.

In fact, $5^{29}=186,264,514,923,095,703,125$ so you are within $10\%$

Answer 2

You are totally correct upto getting $5^{29}=10^{20.3}$ Now just look at data provided: $\log_{10}2 = 0.3$. This translates to:

$10^{0.3}=2$. Thus, $10^{20.3}=10^{20}*10^{0.3}=10^{20}*2$. So, the answer is $21$ digits.

Answer 3

$$\log_{10}\left(\frac{10}2\right)^{29}=29\,(1-\log2)\approx 20.3$$ so your number will have $21$ figures.

Note that the given approximation is sufficient, because $20<29\,(1-0.295)<21$ and $20<29\,(1-0.305)<21$.