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How many $(a,b)$ for $a,b \in \Bbb{N}$ pairs can satisfy the following equation: $$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$ The answer is $3$, but I can't figure out how to get that answer.


This is my attempt.

$$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$ $$\frac{1}{a}\log_2\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$ $$\log_2\left(\log_{2^b}\left(2^{1000}\right)\right)=a$$ $$\log_{2^b}\left(2^{1000}\right)=2^a$$ $$\frac{1}{b}\log_{2}\left(2^{1000}\right)=2^a$$ $$\log_{2}\left(2^{1000}\right)=2^ab$$ $$2^{1000}=2^{2^ab}$$ $$1000=2^ab$$ That's it! This is dead end.

Honestly, this is the best I could do altough I very much doubt that I can get two variables by solving one equation (for that we need a system of equations!). So, I think that I need another approach that will either give me what $a$ and $b$ can be or direct answer (i.e. the number of possible values for $a$ and $b$), but I don't know which one.

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  • $\begingroup$ You probably meant $a,b\in\Bbb{N}$ otherwise we also have $a=-1,b=2000$,$a=-2,b=4000$ etc... $\endgroup$ – kingW3 Jun 12 '18 at 16:26
  • $\begingroup$ Hmmm... probably, but I'll check the original question. $\endgroup$ – Hanlon Jun 12 '18 at 16:33
  • $\begingroup$ Yes. You are right. $\endgroup$ – Hanlon Jun 12 '18 at 16:33
  • $\begingroup$ Depending on the definition of $\mathrm{N}$, we might also have $(a,b)=(0,1000)$. $\endgroup$ – robjohn Jun 12 '18 at 17:22
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I agree with your derivation

$$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1\iff \log_{2^b}\left(2^{1000}\right)=2^a\iff (2^b)^{2^a}=2^{1000}\iff b\cdot 2^a=1000$$

now we can have

  • $a=1, 2^a=2, b=500$

  • $a=2, 2^a=4, b=250$

  • $a=3, 2^a=8, b=125$

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  • $\begingroup$ How did you infer those values from $1000=b2^a$, i.e. how did you know that exactly those values are correct? What is the systematic way to get them? $\endgroup$ – Hanlon Jun 12 '18 at 16:15
  • $\begingroup$ I guess the question doesn't stipulate integers. $\endgroup$ – stuart stevenson Jun 12 '18 at 16:15
  • $\begingroup$ @Hanlon Indeed we can have only three solutions for a and b integers otherwise we woyld have infinitely many solutions. To find that we can start form a=1 and then proceed to find a=2 and a=3. From a=4 we can't have solutions since $b$ is no more integer. $\endgroup$ – gimusi Jun 12 '18 at 16:22

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