Results 1 to 7 of 7

Math Help - An equation with unknown exponent

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    13

    An equation with unknown exponent

    Is is possible to solve this equation?

    0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}

    If so, how?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    several ways come to mind:

    1. Newton's method - Wikipedia, the free encyclopedia

    2. Bisection method - Wikipedia, the free encyclopedia

    3. Secant method - Wikipedia, the free encyclopedia

    4. you can also plot the two functions on each side of the equation on the same system of axis, and find their intersection point.

    5. you can try to converge to the solution:

    <br />
0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}<br />

    <br />
 \Leftrightarrow n_{k + 1}  = \frac{{\ln \left( {0.72 \times 1.04^{n_k }  + 0.28} \right)}}<br />
{{\ln 1.035568}}<br />

    the stopping condition would be: <br />
\left| {n_{k + 1}  - n_k } \right| < \varepsilon <br />

    in this case one obvious solution is n = 0
    Last edited by Peritus; January 28th 2008 at 12:37 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by p.numminen View Post
    Is is possible to solve this equation?

    0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}

    If so, how?
    It would be interesting for us to know where you got your values from.

    If the 1.04 is the rounded value of 1.035568 then we can label this constant as c and your equation becomes:

    0.72 \cdot c^n + 0.28 = c^n~\iff~0.28 = 0.28 \cdot c^n ~\iff~ 1 = c^n~\implies~ n = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2007
    Posts
    13
    No, 1.04 is not the rounded value of 1.035568. They are both exact numbers, actually.

    I made the equation for a practical problem, but then I realised that I cannot solve the equation myself.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by p.numminen View Post
    Is is possible to solve this equation?

    0.72 \times 1.04^{n} + 0.28 = 1.035568^{n}

    If so, how?
    As the others have mentioned, an obvious solution is n=0

    0.72 \times 1.04^{0} + 0.28 = 1.035568^{0}

    0.72  + 0.28 = 1
    Follow Math Help Forum on Facebook and Google+

  6. #6
    dennis
    Guest

    I guess A(B^x) + Cx = D is also a pain?

    I guess A(B^x) + Cx = D is also a pain?

    Capital alphas known constants.

    Guess it requires some transcendental meditation? Trouble is, I've got a few to do, different sets of alphas.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    1

    solve using logs



    the solution to your problem can be worked out using the natural log. I need a brush up on the rules but basically you take the log of all the numbers. The ones that are multipled turn to addition and the exponent comes down....

    (ln 0.72) + (n x ln(1.04)) [not sure how to handle the 0.28] = n x ln 1.035568


    i just dont remember the rules surroundint he 0.28,. i dont think you add it at that point (once you take the ln) but you can look that up.

    I hit this forum because i am looking at stocks with different yeilds and I know their lines cross when graphed, just trying to figure out where. In my case the "time" is the exponent and its easily solved the way i showed here.... EXCEPT.. I am stuck on basic algebra...... once I get to something like T x A + B = T * C [the t is the unknown and abc are known numbers) I completely spaced on how to solve for T. LOL.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Matrix equation with unknown coefficents
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 5th 2011, 10:20 AM
  2. Solve For the Unknown (ln equation)
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 6th 2011, 07:02 PM
  3. 1 unknown equation solver
    Posted in the Math Software Forum
    Replies: 3
    Last Post: June 7th 2011, 01:42 PM
  4. Replies: 7
    Last Post: April 28th 2010, 07:59 AM
  5. Simultaneous equation 2 unknown vairables
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 9th 2009, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum