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#1 windoboy

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Posted 24 February 2010 - 04:18 PM

3^x +4(3^-x)  = 5

solve in R


#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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Posted 24 February 2010 - 05:15 PM

http://www.wolframalpha.com/input/?i=3^x+%...283^-x%29+%3D+5

#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....

#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)

Edited by Jewified, 27 February 2010 - 05:49 PM.





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