# higher maths

genius
Posers and Puzzles 21 May '03 12:43
1. genius
Wayward Soul
21 May '03 12:43
prove that the graphs y=(a^x)+1 and y=a^(x+1) intersect at the point where the x-coordinate is log (1/(a-1)) to the base a, when a&gt;2.

i just sat my higher maths paper today! and this Q was -erm- well, it was the last Q, and no-one that i know got it...ðŸ˜›
2. royalchicken
CHAOS GHOST!!!
21 May '03 19:081 edit
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&gt;=2.
3. genius
Wayward Soul
21 May '03 20:58
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.
the a&gt;2 bit was for the earlier question where we had to draw the graphs :p
4. genius
Wayward Soul
22 May '03 09:31
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˜€
5. royalchicken
CHAOS GHOST!!!
22 May '03 19:51
Erm...I did it too. You WANT something to be to the power of a log.
6. genius
Wayward Soul
22 May '03 21:32
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))
7. royalchicken
CHAOS GHOST!!!
23 May '03 00:36
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.

8. genius
Wayward Soul
23 May '03 08:36
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˜›)
9. royalchicken
CHAOS GHOST!!!
23 May '03 19:161 edit
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.

I posted the solution in that thread.