# Are these two expressions equal to each other?

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 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?

Thanks

• Assume by contradiction that the two numbers are equal, write $13=12+1$ and eliminate fractions. You arrive at the contradiction $12=2^{\ldots}$. – Giuseppe Negro Feb 17 '14 at 23:14
• Can you write out the steps? – omega Feb 17 '14 at 23:17
• 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. – Giuseppe Negro Feb 17 '14 at 23:21
• But if I try to calculate $a$ in the math software, I get $13.4638$..... – omega Feb 17 '14 at 23:24
• Even in floating point, you should expect at least six significant figures, not two (barring catastrophic cancellation, which isn't happening here). – tabstop Feb 17 '14 at 23:38

$$\frac{12}{x}=\frac{13}{1+x}$$ 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 $\,a = 12/37\in\Bbb Q$