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How can I solve this question:

Given that $\log 3$ is about $0.48$, approximately how many digits are in the number $10^{150}$ if it were written in base $3$.

Thanks!

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  • $\begingroup$ Sorry, I'm am new to this website, so I do not know all the formatting tools. Can you tell me how to write exponents without using the "^" symbol? $\endgroup$ – stemrunner123 Jan 8 '16 at 4:48
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Here is a simple solution.

Lets assume that the number of digits in the base 3 representation of $10^{150}$ is $k$.

So, $10^{150}=3^{k-1}a_{k-1}+3^{k-2}a_{k-2}+\cdots+3a_1+a_0$ where $0 \leq a_i \leq 2$ where $a_{k-1}$ is not $0$

Now lets try to put some bounds on the R.H.S. of above.

From the above equation we have $10^{150} \geq 3^{k-1}$.

Also, $3^{k-1}a_{k-1}+3^{k-2}a_{k-2}+\cdots+3a_1+a_0 \leq 2(3^{k-1}+3^{k-2}+\cdots +3+1)=2\left(\frac{3^{k}-1}{3-1} \right) < 3^k$.

So, $3^{k} > 10^{150} \geq 3^{k-1}$.

Since $\log_{3}(x)$ is an increasing function we have,

$k > 150\log_{3}(10) \geq k-1$.

So, $k=1+ \lfloor{150\log_{3}(10)} \rfloor$

The above method of proof clearly points at a general result. What is that?

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Hint: $\log_3 10^{150} = 150\log_3 10 = \dfrac{150}{\log_{10}3} \approx \dfrac{150}{0.48} = 312.5$. This tells you that $10^{150} \approx 3^{312.5}$.

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