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Exponential equation

  1. Jun 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the exponential equation.
    1. (22x)+(2x) - 12 = 0
    2. 3 * 4x + 4 * 2x + 8 = 0

    2. Relevant equations



    3. The attempt at a solution
    1. 2x ln 2 + x ln 2 = ln 12
    x(2 ln 2 + ln 2)= ln 12
    x=( ln 12) / (2 ln 4).
    I'm thinking i'm factoring out the X wrong.

    2. 12 ( 4x * 2x = -8
    4x * 2x = -2/3
    x (ln 8 ) = -2/3
    x = (-2/3) / (ln 8)
    The back of the book says there is no real solution. Why? Lets say it was a real solution. Did i solve the problem correctly?

    Thanks for the help.
    -Scrythe
     
  2. jcsd
  3. Jun 25, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are doing some unorthodox, i.e. wrong, algebra there. It's not true that ln(2^(2x)+2^x)=ln(2^(2x))+ln(2^x) or that 3*4^x+4*2^x=12*(4^x*2^x). The way to solve the first one is to notice that if you put u=2^x then 2^(2x)=u^2. So the equation turns into a quadratic u^2+u-12=0. Solve the quadratic and then figure out x from u. Can you turn the second one into a quadratic as well? Oh, and practice your algebra!!!
     
  4. Jun 25, 2009 #3

    Mark44

    Staff: Mentor

    You have algebra mistakes on almost every single line of your work.
    For problem 1, you have
    As Dick already commented, your first equation is incorrect. The second equation has no error in it. The third equation has an error, because 2 ln 2 + ln 2 != 2 ln 4.
    2 ln 2 + ln 2 = 3 ln 2 = ln 2^3 = ln 8.

    Your work in problem 2 also has some errors.
    Dick already pointed out the error in your first equation above. Going from your first equation to the next, no errors. In the third equation, the error is that 4x * 2x = 8x^2, which is not at all equal to x ln 8.

    Your basic algebra skills are killing your ability to work with more complicated concepts, such as logarithms and exponents. For each of the mistakes that Dick and I pointed out, you should go back and review those topics to make sure that you really understand them. If you don't do that, you're going to have a very tough time with the current topics you're studying.


    equivalent to the first equation, so no mistake.
    The first line above
     
  5. Jun 25, 2009 #4
    it was (4^x) * (2^x) not 4x * 2x, but nevertheless i was able to find that there was no solution because the discriminant was negative.
    (12^1+x) + (8^1+x)
    (u^2) + u + 8 = 0
    (b^2) - 4ac = 1 - 32 = no real solution.

    i was also able to solve the first one. ( though i won't post my work). I realize i am no expert in algebra, but i have not had troubles until this section. I will look back at the algebra techniques.

    thanks again,
    Scrythe
     
  6. Jun 25, 2009 #5

    Mark44

    Staff: Mentor

    Sorry, I just copied your work, and the superscripts got lost in the copy.

    For the second problem, your answer is correct, but your work isn't. The original equation is 3*4x + 4*2x + 8 = 0
    If you make the substitution u = 2x, the equation becomes 3u2 + 4u + 8 = 0. This is quite different from what you posted, which was u2 + u + 8 = 0.

    I have no idea of where you got this:
     
  7. Jun 25, 2009 #6
    Ok that makes sense. I did 3 * 4^x = 12 ^(1+x) and 4 * 2^x = 8^(1+x), which i now realize is incorrect.
     
  8. Jun 25, 2009 #7

    Mark44

    Staff: Mentor

    And that's what we mean by going back over those old topics. As you go further with mathematics, if you don't have the basic stuff down cold, you'll spend all of your time making mechanical errors and won't have a prayer at understanding the higher-level concepts.
     
  9. Jun 26, 2009 #8
    And why don't you substitute y=2x and solve the quadratic equation?
     
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