Im working on a proof, and currently I'm trying to check some expressions to see if they are equal to each other. Using specific values as a test case, I get this expression

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But I can't tell if they are equal to each other.

If I try simplifying them myself, I get $log_2{(12)} = 2log_2{(37)}-7$ and I still don't know if they are equal to each other. The difference of this is about $0.166056$ according to a math software. I think it's because of rounding errors, but I want to be sure.

Does anyone know if there is a way to check?


  • $\begingroup$ Assume by contradiction that the two numbers are equal, write $13=12+1$ and eliminate fractions. You arrive at the contradiction $12=2^{\ldots}$. $\endgroup$ Commented Feb 17, 2014 at 23:14
  • $\begingroup$ Can you write out the steps? $\endgroup$
    – omega
    Commented Feb 17, 2014 at 23:17
  • $\begingroup$ Write $a=2^{7+2\log_2\ldots}$. Then we are asserting by contradiction that $$\frac{12}{a}=\frac{12+1}{1+a}.$$That is, $$12+12a = 12a +a.$$ And so we obtain the contradiction $12=a$: this is clearly impossible since $a$ is a very big number. $\endgroup$ Commented Feb 17, 2014 at 23:21
  • $\begingroup$ But if I try to calculate $a$ in the math software, I get $13.4638$..... $\endgroup$
    – omega
    Commented Feb 17, 2014 at 23:24
  • 1
    $\begingroup$ Even in floating point, you should expect at least six significant figures, not two (barring catastrophic cancellation, which isn't happening here). $\endgroup$
    – tabstop
    Commented Feb 17, 2014 at 23:38

2 Answers 2


$$\frac{12}{x}=\frac{13}{1+x}$$ from here we have $x=12$ then $$2^y=12$$ and we have $y=\log_2 12$ assume

$$7+2(\log_2 12 - \log_237)=\log_2 12$$

is true then you have $$\log_2 128=7 = \log_2 37^2-\log_2 (12)=\log_2 37^2/12$$

since $37^2/12\neq 128$, your assumption is incorrect.


Hint $\ 12 = 2^{\large 7+2\log_2(a)} = 2^7 a^2\,\Rightarrow\, a^2 = 6/2^6\,\Rightarrow\,a = \sqrt{6}/8,\,$ contra $\,a = 12/37\in\Bbb Q$


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