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

Guest Mar 20, 2020

#1**+2 **

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.

jfan17 Mar 21, 2020