Rational Tuple Logarithm Count
How many distinct four-tuples $ (a, b, c, d) $ of rational numbers are there with
\[a \cdot \log_{10} 2+b \cdot \log_{10} 3 +c \cdot \log_{10} 5 + d \cdot \log_{10} 7 = 2005?\]
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We can write the given equation as
\[\log_{10} 2^a + \log_{10} 3^b + \log_{10} 5^c + \log_{10} 7^d = 2005.\]Then
\[\log_{10} (2^a \cdot 3^b \cdot 5^c \cdot 7^d) = 2005,\]so $ 2^a \cdot 3^b \cdot 5^c \cdot 7^d = 10^{2005} $.
Since $ a, $ $ b, $ $ c, $ $ d $ are all rational, there exists some positive integer $ M $ so that $ aM, $ $ bM, $ $ cM, $ $ dM $ are all integers. Then
\[2^{aM} \cdot 3^{bM} \cdot 5^{cM} \cdot 7^{dM} = 10^{2005M} = 2^{2005M} \cdot 5^{2005M}.\]From unique factorization, we must have $ aM = 2005M, $ $ bM = 0, $ $ cM = 2005M, $ and $ dM = 0 $. Then $ a = 2005, $ $ b = 0, $ $ c = 2005, $ and $ d = 0 $. Thus, there is only $ \boxed{1} $ quadruple, namely $ (a,b,c,d) = (2005,0,2005,0) $.
Freeflows