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How many distinct four tuple (a,b,c,d) of rational numbers are there with $a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7=2005$

Can we proceed like this :

Using $\log a +\log b = \log(ab)$ and $m\log a = \log a^m$

$\Rightarrow \log_{10}2^a \cdot 3^b \cdot 5^c \cdot 7^d = 2005$

Please guide how to proceed further..

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2 Answers 2

up vote 3 down vote accepted

You have $$ 2^a\cdot 3^b\cdot 5^c\cdot 7^d = 10^{2005} = 2^{2005}\cdot 5^{2005}. $$ So this works if $a=c=2005$ and $b=d=0$.

To see if other solutions exist, observe that you'd get $$ 2^{a-2005}\cdot 5^{c-2005} = 3^{-b}\cdot 7^{-d}. $$ All the exponents are rational, so we have $$ 2^{n_1/n_2}\cdot 5^{n_3/n_4} = 3^{n_5/n_6}\cdot 7^{n_7/n_8}. $$ Raising both sides to the power $n_2 n_4 n_6 n_8$, we get all integer exponents. By uniqueness of prime factorizations, that can happen only if all of the integer exponenents are $0$. Therefore there can be no other solutions.

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Hint: $10^{2005}=2^a \cdot 3^b \cdot 5^c \cdot 7^d$

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