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03102015  #1  
Join Date: Dec 2011
Posts: 386

Challenging Algebra
When is the following true?
3^x+1=9^x
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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 

03112015  #2  
Join Date: Dec 2008
Posts: 249

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; 03122015 at 09:37 AM.. Reason: Solution to problem 

03172015  #3  
Join Date: Dec 2011
Posts: 386

I apologize for leaving out the paretheses i made you do harder work But your answer is impeccable.
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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 

05202015  #4  
Join Date: May 2015
Posts: 5

It's 1 with the parentheses and 0.43801787946 without them.


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