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I have in front of me a math problem that I do not understand. That's to say, I don't understand what is being asked of me.


We can define $\log_2**(x) = log_2*(log_2*(x))$ and the function $log_2**...**(n-times)$ as the function you get when the log* function is put together n times (like above). Let $\beta(x)=min\{k|log_2**...**(k-times)(x)\leq2\}.$ Solve the equation $\beta(x)=3$

At first, I thought I might me able to put $k=1$ or $k=0$ and $x=8$ making it $log_2(8)=3$ but apparently, that is not allowed.

Anyone get what is being asked here?

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up vote 2 down vote accepted

What is being asked of you is "what is the minimum number of times must you take log_2 of x to get a number less than or equal to 2"

To give some examples of what this means

$\beta(2) = 0$ as you need not take it's log to get it $\leq 2$

$\beta(3) = 1$ as $\log_2(3) < 2$

$\beta(2^{16}) = 3$ as $\log_2(2^{16}) = 16$, $\log_2(16)=4$, $\log_2(4)=2$

does that make more sense?

If we want to find the minimum $x$ with $\beta(x)=3$ we would note that we need 2^2^k for any $2\geq k>1$

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Yes, this does help. Thanks! – Mappan Oct 3 '13 at 1:00
Do note that I just revised it, if k>2, then there are problems again, you really want k between 1 and 2 – Vernepator Cur Oct 3 '13 at 1:01

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