 Intermediate Challenging Algebra - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Intermediate Challenging Algebra 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.  Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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