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.

  • $\begingroup$ 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

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

  • $a=1, 2^a=2, b=500$

  • $a=2, 2^a=4, b=250$

  • $a=3, 2^a=8, b=125$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.