Thread: log problem

Please find attached problem and my work.

Originally Posted by Veronica1999

Please find attached problem and my work.

The integers with rational logs to the base 8 are all of the form $\displaystyle 2^k$ for some natural number $\displaystyle $$k$.

CB

How do I prove that log(base8)6 has an irrational value?
I can split this into log(base8)2 + log(base8)3 = 1/3 + ?

Originally Posted by Veronica1999

How do I prove that log(base8)6 has an irrational value?
I can split this into log(base8)2 + log(base8)3 = 1/3 + ?

Proof by contradiction. Assume it is rational.

$\displaystyle \log_{8} 6 = \frac{a}{b}$

$\displaystyle 8^{\frac{a}{b}} = 6$

$\displaystyle 8^a = 6^b$

$\displaystyle 2^{3a} = 2^b*3^b$

$\displaystyle 2^{3a-b} = 3^b$

From unique factorization, this is impossible - a contradiction.

QED.