Rational Logarithm Solutions
How many distinct four-tuples $ (a,b,c,d) $ of rational numbers are there with \[a\log_{10} 2+b\log_{10} 3+c\log_{10} 5+d\log_{10} 7=2005?\]
- 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
The given equation is equivalent to \[
\log_{10}\left(2^a\cdot 3^b\cdot 5^c\cdot 7^d\right)= 2005, \quad \text{so} \quad 2^a\cdot 3^b\cdot 5^c\cdot 7^d = 10^{2005} = 2^{2005}\cdot 5^{2005}.\]Express $ a $, $ b $, $ c $ and $ d $ as simplified fractions and let $ M $ be the least common multiple of their denominators. It follows that \[
2^{Ma}\cdot 3^{Mb}\cdot 5^{Mc}\cdot 7^{Md} = 2^{2005M}\cdot 5^{2005M}.\]Since the exponents are all integers, the Fundamental Theorem of Arithmetic implies that \[
Ma = 2005M, \quad Mb = 0, \quad Mc= 2005M,\quad\text{and}\quad Md = 0.\]Hence the only solution is $ (a, b, c, d) = ( 2005, 0, 2005, 0) $, and the answer is $ \boxed{1} $.
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