Is $\log_27$ a rational number?

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    $\begingroup$ Hint: suppose it were, so that $\log_2 7=\frac ab$ for some natural numbers $a,b$. What could you deduce? $\endgroup$ – lulu May 23 '16 at 15:07
  • $\begingroup$ I deduced that ,suppose a=3, then 8^1/b=7,but b cannot be natural . $\endgroup$ – prashik May 23 '16 at 15:13
  • $\begingroup$ Well...you won't get far eliminating integers one at a time. What can you deduce for an unspecified $a,b$? $\endgroup$ – lulu May 23 '16 at 15:16

Suppose $\log_2 7 = {a\over b}$ for two positive integers $a$ and $b$.

$\log_2 7 = { \ln 7 \over \ln 2} = {a \over b}$

Cross multiply,

$b \ln 7 = a \ln 2 \implies \ln ( 7^b ) = \ln (2^a)$

Take the $e^{( \ \ )}$ of both sides,

$7^b = 2^a$

This is impossible for integers $a$ and $b$ because $7^b$ is always going to be an odd number, while $2^a$ will always be an even number. They can never be equal, thus, $\log_2 7$ is not a rational number.

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    $\begingroup$ Or: If $\log_{2} 7 = \frac{a}{b}$, then $7 = 2^{a/b}$ by the definition of a logarithm, so $7^{b} = 2^{a}$, etc. :) $\endgroup$ – Andrew D. Hwang May 23 '16 at 15:28

If $a,b$ are positive integers and $2^{a/b}=7$ then $2^a=7^b$ which make an even number equal to an odd number.


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