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I would like to know are the some $p \in \mathbb{N}$ and $q \in\mathbb{N}$ for $3^p = 4^q$ except the trivial $p = q = 0$. So, I entered the expression into Wolfram Alpha, which returned the result for $q$:

$\displaystyle q = \frac{\log(3^p)}{\log(4)} + \frac{2\pi i c_{1}}{\log(4)}$

Nothing is said about what is $c_{1}$. Arbitrary constant? If $c_{1} \neq 0$ the result could be complex-valued I guess. I tried some calculations using the formula but the results do not make sense.

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user171297
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    The denominator should probably be $\log 4$ instead of $\log 3$. Anyway, that $c_1$ is supposed to be an integer parameter. Mathematica's messages are, perhaps, more informative. Well, I paid good money for it :-) – Jyrki Lahtonen Aug 23 '14 at 19:08
  • You are right. I corrected the error. – user171297 Aug 23 '14 at 19:09
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    Also the fundamental theorem of arithmetic tells immediately that $p=q=0$ is the only solution with $p,q$ integers. – Jyrki Lahtonen Aug 23 '14 at 19:10
  • $c_1$ must be an integer. In general, since $e^{2\pi i}=1$, complex logarithms are multi-valued. So $4^{2\pi i/\log 4}=1$. – Thomas Andrews Aug 23 '14 at 19:10
  • But Wolfram Alpha is not going to restrict to natural numbers, so that formula is almost certainly not useful, except, if $p,q$ are natural numbers, then you can say $c_1=0$. The resulting formula won't help, though. – Thomas Andrews Aug 23 '14 at 19:13
  • @ThomasAndrews It will (or will try to) if you tell it to. I tried http://www.wolframalpha.com/input/?i=solve+3%5Ep+%3D+4%5Eq+in+integers –  Aug 25 '14 at 14:32
  • For some idea of a baseline, try the query "solve 2^p = 4^q in integers". –  Aug 25 '14 at 14:54

2 Answers2

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Other's have discussed the output given by Wolfram|Alpha, so I'll provide only a hint for a solution.

Hint: For every $p \in \mathbb N$ with $p\geq 1$, we have that $4^p$ is even but $3^p$ is odd. A natural number cannot be both even and odd, so...

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So to find all solutions, real and complex consider the definition of an exponential. $a^b=e^{b\log a}$. Also helps to mention that the complex logarithm of a complex number $r e^{it}$ is $\log\mid r \mid+2n\pi i$ with $n$ an integer. It's to do with the fact that the logarithm function is the inverse of a function that isn't one to one so you've to choose a 'branch'. Anyway...

Now write as $ e^{p\log 3} = e^{4\log q} $.

Take the complex log of both sides so that

$p\log 3 + 2n\pi i=q\log 4$ where $n$ is an integer and then divide across

snulty
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