log problem

**Veronica1999** · November 30th 2010, 10:28 PM

Please find attached problem and my work.

**CaptainBlack** · December 1st 2010, 12:12 AM

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 $2^k$ for some natural number $$$k$ .

CB

**Veronica1999** · December 1st 2010, 01:23 PM

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 + ?

**elemental** · December 1st 2010, 02:27 PM

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.

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

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

$8^a = 6^b$

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

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

From unique factorization, this is impossible - a contradiction.

QED.

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