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  • Need help in simplifying this problem

    Here's a problem I was working with(MFE). I understand the concepts of it(so won't bother to post the question in its entirety). However, I am having problems simplifying it to find x.

    As you can see, both sides of the "=" sign have exponentials with the x(which we want to find) in it. I forgot how to do so to find x. Could someone show me step by step?

    21.4949 + 119e^(-1.5x) = 9.1927 + 129 - 8.06e^(-.25x)

    Thanks!

  • #2
    Originally posted by StatGuy3 View Post
    Here's a problem I was working with(MFE). I understand the concepts of it(so won't bother to post the question in its entirety). However, I am having problems simplifying it to find x.

    As you can see, both sides of the "=" sign have exponentials with the x(which we want to find) in it. I forgot how to do so to find x. Could someone show me step by step?

    21.4949 + 119e^(-1.5x) = 9.1927 + 129 - 8.06e^(-.25x)

    Thanks!
    Looks like you'd have to use an approx. technique assuming you did everything right...

    Comment


    • #3
      119 exp(-1.5x)+8.06 exp(-.25x)=116.6978
      ln(119 exp(-1.5x)+8.06 exp(-.25x))~4.76

      (ln(119)-1.5x)*(ln(8.06)-.25x))~4.76
      a quadratic appears, solve it for x
      Cheers !

      Comment


      • #4
        Originally posted by shima View Post
        119 exp(-1.5x)+8.06 exp(-.25x)=116.6978
        ln(119 exp(-1.5x)+8.06 exp(-.25x))~4.76

        (ln(119)-1.5x)*(ln(8.06)-.25x))~4.76
        a quadratic appears, solve it for x
        Try again, ln(a+b) is not ln(a*b). However ln(a*b) = ln(a) + ln(b)

        Comment


        • #5
          There should be a "nice way" to solve the equation if this is a textbook/practice exam problem. The CAS/SOA do not give us questions that require numerical methods. However, since I love math, I will pass on a very simple and neat tool from my toolbox that you can use to solve this.

          If you want a simple approximation method, Newton's Method is easily done with the TI-30X, then try this recursive formula. If x is supposed to be an interest rate, then I recommend starting with n=0.5
          1.) Put all terms on one side
          2.) Let f.1 = derivative of f(X)
          3.) X(n+1)=X(n) - f(n)/f.1(n)
          4.) Use X(n+1) as the input for the right hand side again, so you will then find X(n+2)=X(n+1) - f(n+1)/f.1(n+1)
          Usually, you will have to cycle through your A-E memory registers no more than 2 times. Usually by the second time you get to E again, the numbers have converged to one of the answer choices.
          Dan Kamka
          P/1, FM/2

          Comment


          • #6
            Originally posted by dkamka View Post
            There should be a "nice way" to solve the equation if this is a textbook/practice exam problem. The CAS/SOA do not give us questions that require numerical methods. However, since I love math, I will pass on a very simple and neat tool from my toolbox that you can use to solve this.

            If you want a simple approximation method, Newton's Method is easily done with the TI-30X, then try this recursive formula. If x is supposed to be an interest rate, then I recommend starting with n=0.5
            1.) Put all terms on one side
            2.) Let f.1 = derivative of f(X)
            3.) X(n+1)=X(n) - f(n)/f.1(n)
            4.) Use X(n+1) as the input for the right hand side again, so you will then find X(n+2)=X(n+1) - f(n+1)/f.1(n+1)
            Usually, you will have to cycle through your A-E memory registers no more than 2 times. Usually by the second time you get to E again, the numbers have converged to one of the answer choices.
            SOA has given exam problems where in the final step you had to plug in the answers to see which one was right and could not be solved by elementary algebra tricks.

            Comment


            • #7
              I understand now which type of question this is, so it sound like one of those which require you to do some checks and hope for a lucky guess so you don't have to try all 5 possibilities.

              By the way, Newton's Method is not an 'elementary algebra trick'.
              Dan Kamka
              P/1, FM/2

              Comment


              • #8
                Originally posted by dkamka View Post
                I understand now which type of question this is, so it sound like one of those which require you to do some checks and hope for a lucky guess so you don't have to try all 5 possibilities.

                By the way, Newton's Method is not an 'elementary algebra trick'.
                I did not say it was. Actually I said the opposite.

                Comment

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