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Solving Exponential Equations SO CONFUSED ?

  1. Oct 12, 2009 #1

    Monocerotis

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    Solving Exponential Equations SO CONFUSED !?!

    1. The problem statement, all variables and given/known data

    I'm trying to solve problems like these
    a) 2(3y-2) = 18
    b) 27(33x+1) = 3
    c) 2x+2 - 2x = 48
    d) 10z+4 + 10z+3 = 11

    2. Relevant equations

    Laws of exponential values.


    3. The attempt at a solution
    b) 2(3y-2) = 18
    6y-2 = 18
    Stuck/Confused.

    Maybe there's something fundamental I'm forgetting or making overly complicated. If you divide 18 by 3 to make it 6 so then you could just work with the exponential values to solve for y, wouldn't you also have to do the same thing to 2(3y-2) (or) 6y-2

    For say a problem like this
    3(5x+1) =15
    15x+1 = 15

    Now you have both bases balanced, you can work to solve for x
    x+1 = 0 ?
    x = 0 - 1
    x = -1

    That would then mean 150 = 15, which doesn't work out because anything to an exponential value of 0 is 1, correct ?
     
  2. jcsd
  3. Oct 12, 2009 #2
    Re: Solving Exponential Equations SO CONFUSED !?!

    use natural logs to solve for the exponents!
    ie 2(3^y-2) = 18
    3^y-2=9
    ln3^y=ln9
    yln3=ln9
    y=ln9/ln3

    all of them can be solved that way!
     
  4. Oct 12, 2009 #3

    Monocerotis

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    Re: Solving Exponential Equations SO CONFUSED !?!

    Ok that sounds interesting. We haven't learned how to use natural logs in class yet, they begin teaching you how to use that in the next level course.

    Is there any other way to solve the problem without using natural logs ?

    If not I guess I'll just have to teach myself lol
     
  5. Oct 12, 2009 #4
    Re: Solving Exponential Equations SO CONFUSED !?!

    correction!

    2(3^y-2) = 18
    3^y-2=9
    ln3^y-2=ln9
    (y-2)ln3=ln9
    y=(ln9)+2/ln3
     
  6. Oct 12, 2009 #5
    Re: Solving Exponential Equations SO CONFUSED !?!

    Okay! well, if the exponents have the same base (ie 2^x-1=2^6-x) then you can set the exponents equal to each other. For your problems, you will have to manipulate the bases to get them to be the same... for example if you have 2^x-4=8, that also equals 2^x-4=2^3, so then x-4=3.
     
  7. Oct 12, 2009 #6
    Re: Solving Exponential Equations SO CONFUSED !?!

    Also-Because there are exponents involved, don't multiply the number outside of the parentheses inside of the parentheses!
     
  8. Oct 12, 2009 #7
    Re: Solving Exponential Equations SO CONFUSED !?!

    Using natural logs is quite easy (it's just another logarithm, with base e), but you can also try re-writing the side of the equation that does not have exponents:

    2(3y-2) = 18

    3y-2 = 9

    3y-2 = 32

    Once both sides have the same base (3 in this case), you can set the exponents equal to solve for the variable:

    y-2 = 2

    The only problem is that the numbers don't always work out so easily (for example, if you had = 5 instead of = 18, you would need a calculator and end up with horrible decimal numbers).
     
  9. Oct 12, 2009 #8
    Re: Solving Exponential Equations SO CONFUSED !?!

    Excellent explanation!
     
  10. Oct 12, 2009 #9
    Re: Solving Exponential Equations SO CONFUSED !?!

    If you need to take the log of both sides, you can do it in any base (not just ln or loge), such as log10 or log3
     
  11. Oct 12, 2009 #10

    Monocerotis

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    Re: Solving Exponential Equations SO CONFUSED !?!

    Thanks Bohrok, maladroit, and darkchild for your help.

    Don't know what it is but sometimes I just get stuck on problems and can't work my way out of them.

    I think i'll be able to figure it all out now.
     
  12. Oct 12, 2009 #11
    Re: Solving Exponential Equations SO CONFUSED !?!

    All of these problems can be done without the use of logarithms.

    You'll need to be able to factor correctly in the last two to solve that without a calculator.
     
  13. Oct 12, 2009 #12

    Monocerotis

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    Re: Solving Exponential Equations SO CONFUSED !?!

    For this example problem that I listed, I know off the bat the answer need to be x = 0
    so why when I work it out do I end up with x = -1. What am I doing wrong in the problem solving process ?

    3(5x+1) =15
    15x+1 = 15
    --------------------
    x+1 = 0 ?
    x = 0 - 1
    x = -1
     
    Last edited: Oct 12, 2009
  14. Oct 12, 2009 #13

    Mark44

    Staff: Mentor

    Re: Solving Exponential Equations SO CONFUSED !?!

    3(5x + 1) != 15x + 1.
    Hint: Distributive law.
     
  15. Oct 12, 2009 #14
    Re: Solving Exponential Equations SO CONFUSED !?!

    If this is the problem you're referring to:

    3(5^(x+1)) =15
    15^(x+1) = 15

    You can not distribute 3 to 5 since they have different exponents. (1 versus x+1)

    Instead of distributing, try dividing 3 to both sides. The 15 will become a 5 and you can set the exponents on both sides equal.
     
  16. Oct 12, 2009 #15

    Monocerotis

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    Re: Solving Exponential Equations SO CONFUSED !?!

    I think you may be working of my typo, the x+1 should have been x+1 (exponential values). So when 15 x+1 = 15 you "ignore" the values of 15 and work to solve for x with the exponential values.

    So is my problem then neglecting that 15 = 151, so then I would have to work with x+1 = 1
    -> x = 1-1 = 0
     
  17. Oct 12, 2009 #16
    Re: Solving Exponential Equations SO CONFUSED !?!

    Don't distribute!!! Divide both sides by 3, leaving you with 5^x+1=5
    and remember that 5=5^1
     
  18. Oct 12, 2009 #17

    Monocerotis

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    Re: Solving Exponential Equations SO CONFUSED !?!

    thought so

    /facepalm
     
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