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

Mar 20, 2020

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Hey there guest! I'm doing this at 12:30 rn so I didn't get to finish the whole question. I will share what I've gotten thru so far:

Let's first start off with some basic logarithm properties: first,

$$c\log_{a}b = \log_{a}b^c$$

Next:

$$\log_{a}b + \log_{a}c = \log_{a}b+c$$

With these 2 properties, we have enough to finish our question!

For the first term, a * log 10 of 2, we can rewrite that as:

$$\log_{10}2^a$$

Repeating this for all the terms in the given equation, we eventually reach:

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

This simplifies to:

$$\log_{10}(2^a*3^b*5^c*7^d) = 2005$$

Using the basic definition of a log, we can rewrite this as:

$$10^{2005} = (2^a*3^b*5^c*7^d)$$

That's about as far as I got! :/ sorry for the inconvenience!

Now adding on to what I did yesterday:

As far as we're concerned, this problem is now just asking us for which of the prime factor "four-tuples"(in 2, 3, 5, and 7) multiply to equal $$10^{2005}$$. We clearly see that there is only 1 distinct tuple that gives us $$10^{2005}$$, because 10 factors out into $$(2*5)^{2005}$$, meaning that a must be 2005, c must be 2005, and b and d are 0.

Mar 21, 2020
edited by jfan17  Mar 21, 2020
edited by jfan17  Mar 21, 2020