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 April 18th, 2014, 07:43 AM #1 Newbie   Joined: Feb 2012 Posts: 8 Thanks: 0 Solve for y It's probably obvious but not to me. 2^(y+1) = 3*5^y I thought ((Y+1)/y)ln2/ln5)=ln3 would have given me the answer but doesn't seem to check. Cheers.
April 18th, 2014, 07:49 AM   #2
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Quote:
 Originally Posted by Devil#20 It's probably obvious but not to me. 2^(y+1) = 3*5^y I thought ((Y+1)/y)ln2/ln5)=ln3 would have given me the answer but doesn't seem to check. Cheers.
y=(log(3)-log(2))/(log(2)-log(5))

 April 18th, 2014, 08:11 AM #3 Newbie   Joined: Feb 2012 Posts: 8 Thanks: 0 Thanks. Your solution gives the right answer on the check ie. -.4425 but I can't see what steps you used to get your solution.
 April 18th, 2014, 08:18 AM #4 Newbie   Joined: Feb 2012 Posts: 8 Thanks: 0 Ah. The penny has just dropped.
April 18th, 2014, 12:12 PM   #5
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Quote:
 Originally Posted by Devil#20 2^(y+1) = 3*5^y
Tattoo this on your wrist: if a^p = x, then p = log(x) / log(a)

I find it easier/shorter then to go this way:
2^(y + 1) = 2^y 2^1 = 2(2^y)
So:
2(2^y) = 3(5^y)
2^y / 5^y = 3/2
(2/5)^y = 3/2
...and for the grand finale:
y = log(3/2) / log(2/5)

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