Prove that log 9 </msub> &#x2061;<!-- ⁡ --> 15 is irrational Im having tr

Kellen Perkins

Kellen Perkins

Answered question

2022-05-23

Prove that log 9 15 is irrational
Im having trouble with the following proof... Ill post what I have completed so far..
Ill attempt by contradiction assuming log 9 15 is rational.
So,
log 9 15 = a b
15 = 9 a b
15 b = 9 a (This is where I'm getting stuck)
Any hints/tips/advice would be great. Thanks

Answer & Explanation

Conor Frederick

Conor Frederick

Beginner2022-05-24Added 7 answers

a and b are positive integers. 15 b and 9 a are positive integers. 5 definitely does not divide 9 a , so what must b be?
Gael Gardner

Gael Gardner

Beginner2022-05-25Added 4 answers

15 b = 9 a
After this, you can write it as
15 b 9 a = 1 3 b 5 b 3 2 a = 1 5 b 3 2 a b = 1
To fulfill this equation, numerator should be equal to denominator but no power of 3 and 5 will be same because power of 3 will end with 1,3,7,9 and that of 5 will end with 5 only, so their power can only be 0.
2 b = 0 and 2 a b = 0
so b equals zero but it is denominator in our first assumption of rationality, which contradicts our assumption. Therefore, log 9 15 is irrational.

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