3^x +4(3^-x) = 5
solve in R
0
[math] log
Started by windoboy, Feb 24 2010 04:18 PM
5 replies to this topic
#1
windoboy
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Posted 24 February 2010 - 04:18 PM
#2
pezzaperry
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Posted 24 February 2010 - 04:26 PM
woah
#3
windoboy
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Posted 24 February 2010 - 04:30 PM
this is 1 little sentence in a 6 part question in the sub questions within the folds of the lesson i still have the ones after the lesson and at the end of the book (now those are like french cheese....the ugly bitter one)
#4
hfswjyr
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#5
windoboy
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Posted 24 February 2010 - 05:34 PM
thx a lot
but it can't help me as it doesn't show me the way of solving
it gives the answer...
interesting link thou....
but it can't help me as it doesn't show me the way of solving
it gives the answer...
interesting link thou....
#6
Jewified
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Posted 27 February 2010 - 05:48 PM
3^x +4(3^-x) = 5
Multiply by 3^x:
3^2x + 4*3^(-x+x) -5(3^x) = 0
3^2x -5(3^x) + 4 = 0
Now let t = 3^x, so:
t^2 -5t + 4 = 0
Solve for t, substitute the 3^x back and eliminate bad solutions (remember that 3^x>0 for all x in R)
Multiply by 3^x:
3^2x + 4*3^(-x+x) -5(3^x) = 0
3^2x -5(3^x) + 4 = 0
Now let t = 3^x, so:
t^2 -5t + 4 = 0
Solve for t, substitute the 3^x back and eliminate bad solutions (remember that 3^x>0 for all x in R)
Edited by Jewified, 27 February 2010 - 05:49 PM.
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