Originally posted by royalchicken Plug in log (1/(a-1)) into both. They come out equal. QED.

Clearly, base 1 logarithms are undefined, and if a=2, log(1/(a-1)) = 0, so 2^0+1=2 and 2^(0+1) = 2. So you should have said a>=2.

also-you have two variables, x and a, so i'm not sure how that would work out, and i have no idea what to do it something is to the power of a log... what we were meant to do is let one equal the other, and and find out what x equals...i didn't do it correctly in the exam, but i did it at home! ðŸ˜€

Originally posted by royalchicken Erm...I did it too. You WANT something to be to the power of a log.

blah-my teacher just skimmed over the sections on logs...ðŸ˜› why do you want something to be to the power of a log? and what was your answer? mine's is
a^(x+1)=a^x +1
a^(x+1)/a^x=1+a^-x
a^1=1+a^-x
a=1+a^-x
log(a-1)=-xloga
x=log(1/(a-1))

Originally posted by royalchicken All logs to base a.

a^(x+1) = a^x + 1
a - 1 = a^-x
log(a-1) = -x

x = -log(a-1) = log [(a-1)^-1] = log (1/(a-1))

Pretty much the same.

The raising to powers bit is just if you substitute the required x-value directly.

by the way-i did a search for all posts bt alcoyte containing universtity (i was trying to find out if he had said that tum ucker went to his university) and i found a post congratulating you on your solving of the cambridge dons probelm-did you really sove it?!? (i gave up reading that thread cause it was getting to large ðŸ˜›)

I did solve it...and I sent my solution to a professor of Acolyte's. He told me that he had never solved it, but a colleague of his had, and my solution was determined to be correct. There is nothing particularly elegant about my solution, though. The only part of it that could be called 'clever' would be identifying 6 different cases that could occur in the conversations, and proving it for each of them.

I recommend giving it another go. It's an eight-tea problem.