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Please try this question from a math test. I am baffled by it. 4fccab66e4b0c6963ad793df-smishra-1338813325025-copyofnewmicrosoftofficeworddocument.docx

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i dont think it uploaded properly try again and ill try and help
i'll draw it
ok :)

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If a, b, c are positive and are pth, qth and rth terms of a GP; then show without expanding that | loga p 1 | | logb q 1 | = 0 | logc r 1 |
sorry i didnt read
i think its a determinant
what does GP stand for?
geometric series
ah nice. ok give me a minute
i assume that they are consecutive terms - can we do that?
unfortunately not, but we do know their positions, p,q,r
the nth term of a GP is A(r^(n-1)) where A is the first term and r is the common ratio
to avoid confusion lets use d instead of r
so lets call the first term of the progression "f" and the common ratio "d"
can you see this: \[a= fd^{p-1}\]
cool have a go playing with that, and see how far you can get it
b = fd^(q-1)
that was a lovely problem :) made me very happy. i haven't really come across logarithms intertwined with matrices before. thanks for showing me
hows it going?
i'm struggling - whats confusing me is that it says do not expand
ah ok well ill put up my solution and you can read through as far as you like
\[\left|\begin{matrix}log(a) & p & 1\\ log(b) & q & 1\\ log(c) & r & 1\end{matrix}\right| \]
\[\left|\begin{matrix}\log(fd^{p-1}) & p & 1\\ \log(fd^{q-1}) & q & 1\\ \log(fd^{r-1}) & r & 1\end{matrix}\right|\]
now just to check, are you familiar with how you can manipulate determinants, eg row operations?
yes - i got that far! - no more!
only vaguely
thats why I'm stuck
lets start by applying some log laws first we break apart the logarithms like this: \[\left|\begin{matrix}\log(d^{p-1}) + \log(f) & p & 1\\ \log(d^{q-1}) + \log(f) & q & 1\\ \log(d^{r-1}) + \log(f) & r & 1\end{matrix}\right|\]
then: \[\left|\begin{matrix}\ (p-1)\log(d) + \log(f) & p & 1\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
yea i follow that
ok, now we do some row operations. remember row operations do not change the value of the determinant so we can use them here. first we do R1 - R2 ( that's row 1 - row 2) you should get this: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-1)\log(d) + \log(f) & q & 1\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
now we do R2- R3 : \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\]
oh yes i remember doing something like this - to find inverse of matrix i think
i think i know whats coming - when you calculate the determinant the zeros will be significant
probably because you need determinants to find inverses. anyway back to the determinant :D now can you see IF we were to expand the determinant using column 3 to expand, the two left elements of R3 would not contribute to the determinant? yeah you guessed it, but we are not allowed to expand, we'll have to keep manipulating.
because of the two existing zeros we can do this: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ 0 & 0 & 1\end{matrix}\right|\] do you understand or should i explain this step more?
to be honest no
ok no worries, ill explain :)
yes i think i see now - those elements in row 3 are multiplied by 0
how about trying to do the operation (column 2) - r(column 3) here: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & r & 1\end{matrix}\right|\] we would get: \[\left|\begin{matrix}\ (p-q)\log(d) & (p-q)-0 & 0\\ \ (q-r)\log(d) & (q-r) - 0 & 0\\ \ (r-1)\log(d) + \log(f) & (r)- r & 1\end{matrix}\right|\] to get: \[\left|\begin{matrix}\ (p-q)\log(d) & p-q & 0\\ \ (q-r)\log(d) & q-r & 0\\ \ (r-1)\log(d) + \log(f) & 0 & 1\end{matrix}\right|\]
there are two ways to think about it, by expanding or by operations, whichever you prefer :)
if we were to expand it those two elements wouldn't count because, as you said, they just get multiplied by zero
does this help or is it still confusing? i hope i haven't made it worse lol
i think i'm getting it - i remember finding this stuff difficult at a-level and it never 'stuck' in my brain very well . i'll keep at it though - thanks very much - i'm glad you enjoyed doing the problem
basically its all about using row operations. they through you a curve-ball with the logarithms and the GP
here is a good site
thanks a lot
no problem :) thanks for the juicy problem

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