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Exponential Equations

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve.
    Q 1.) [tex]3^{3x}+3^{3x+2}=30[/tex]

    Q 2.) [tex]3^{2x}-12(3^{x})+27=0[/tex]

    2. Relevant equations

    [tex](a^{n}){b}=a^{nb}[/tex]

    [tex]a^{n}(a^{b})=a^{n+b}[/tex]

    3. The attempt at a solution

    Q 1.) [tex]3^{3x}+3^{3x+2}=3^{1}+3^{3}[/tex]

    Q 2.) [tex]3^{2x}+3^{3}=12(3^{x})[/tex]
    [tex]3^{2x}+3^{3}=(3^{1}+3^{2})(3^{x})[/tex]
    [tex]3^{2x}+3^{3}=3{1+x}+3^{2+x}[/tex]
     
  2. jcsd
  3. Sep 16, 2008 #2

    Defennder

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    Homework Helper

    Don't see how that helps. Solve for 3^(3x) first. Then use logarithms to find x.

    Same technique as above. This time since it's a quadratic expression you have solve for 3^x using the quadratic formula. Then apply logs again.
     
  4. Sep 16, 2008 #3
    Ignore on Question 2 my last proof of work, it's wrong, its supposed to be:

    [tex]3^{2x}+3^{3}=3^{1+x}+3^{2+x}[/tex]

    And I would use logs but this question can supposedly be done without logs (the teacher hasn't shown us how to do them yet).
     
  5. Sep 16, 2008 #4
    Hint

    Q1. Let [tex]y = 3^{3x}[/tex]. Can you write equation in terms of y?
     
  6. Sep 16, 2008 #5
    Alrighty.

    1. You can use the rule that states that ab * an = ab+n
    33x + 33x+2=30
    33x + 33x*32=30
    Now you can isolate the 33x on the left hand side:
    33x (1 + 9) = 30
    33x=31
    3x=1
    x=1/3

    Sorry, I'm not too sure about #2 yet. I'll keep trying it and I'll keep you posted if my mind comes up with something. :)
     
  7. Sep 16, 2008 #6
    How did you go to this:
    33x (1 + 9) = 30

    I don't get how that works, can you explain it?
     
  8. Sep 16, 2008 #7
    So you mean I'd have like [tex]3^{6x}(3^{2})=30[/tex]?

    EDIT: Nevermind I understand how you did it now, Same with my previous post I understand it, I'll work on the other question. Thanks.

    By the way, Defennder, how did you know it was a quadratic expression?
     
    Last edited: Sep 16, 2008
  9. Sep 16, 2008 #8
    Yeah, of course.

    33x+33x*32=30
    You see how both terms on the left have 33x in them, right? So you can factor out that part. That will leave you 1 instead of the first term, and 32 instead of the second.
    33x*(1+32)=30
    32 is 9... Does that help at all?
     
  10. Sep 16, 2008 #9
    Ya that makes sense, thanks. I just can't find any like terms in the second question now.
     
  11. Sep 16, 2008 #10
    Ok I figured out the last question, and why it was in a quadratic, sorry for all the posts. Here's my solution:

    [tex]3^{2x}-12(3^{x})+27=0[/tex]
    Like mentioned earlier, a quadratic:
    [tex]x^{2}-12x+27[/tex]
    [tex](x-9)(x-3)=0[/tex]
    [tex](3^{x}-3^{2})(3^{x}-3^{1})[/tex]
    x=2, x=1
     
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