Is log 2 base 10 an irrational number

Assume log 2 is a rational no, i.e: log 2=p/q, wherep,q are integers and q=/=0

log2=p/q
q log 2 = p
2^q=10^q=2(5^q)
2^(q-1)=5^q
contradiction arised as 2^(q-1) is always even number and 5^q is always odd.
i.e. log 2 base 10 an irrational number

2008-06-18 22:56:15 補充：
2^q=10^p=(2x5)^p
2^(q-p)=5^p

ok, I got it wrong

Some changes on Lee Sheung Pui's answer.

Assume log 2 is a rational no, i.e: log 2=p/q, where p,q are integers and q=/=0

log2=p/q
q log 2 = p
2^q=10^p=(2x5)^p
2^(q-p)=5^p
contradiction arised as 2^(q-p) is always even number and 5^p is always odd.
i.e. log 2 base 10 an irrational number