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 03-10-2015 #1 Dragonballful23     Join Date: Dec 2011 Posts: 386 Challenging Algebra When is the following true? 3^x+1=9^x __________________ Finally finished all my tests!!! I am ekko main now ^_^ (ign: supersaiyan2363) It's not how much time you have, it's how you use it. -Ekko
03-11-2015   #2
MAS1

Join Date: Dec 2008
Posts: 249

Quote:
 Originally Posted by Dragonballful23 When is the following true? 3^x+1=9^x
Did you mean 3^(x+1) = 9^x?

If so, then:

3^(x+1) = (3^2)^x
3^(x+1) = 3^(2x)
x+1 = 2x
x = 1

But the way the problem is originally written is much harder to solve.

(3^x) + 1 = 9^x
(3^x) + 1 = 3^(2x)
1 = 3^(2x) - 3^x
3^(2x) - 3^x - 1 = 0

Let a = 3^x

a^2 - a - 1 = 0
Then using the quadratic formula a = (1 + sqrt(5))/2 which is the golden ratio!

So 3^x = (1 + sqrt(5))/2
Taking log base 3 (log3) of both sides gives:
x = log3[(1 + sqrt(5))/2]

or using natural logs x = [ln(1 + sqrt(5)) - ln(2)]/(ln(3))

Last edited by MAS1; 03-12-2015 at 09:37 AM.. Reason: Solution to problem

 03-17-2015 #3 Dragonballful23     Join Date: Dec 2011 Posts: 386 I apologize for leaving out the paretheses i made you do harder work But your answer is impeccable. __________________ Finally finished all my tests!!! I am ekko main now ^_^ (ign: supersaiyan2363) It's not how much time you have, it's how you use it. -Ekko
 05-20-2015 #4 eliseo   Join Date: May 2015 Posts: 5 It's 1 with the parentheses and 0.43801787946 without them.

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