Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)
-   -   Exponential Equations (https://www.askmehelpdesk.com/showthread.php?t=347298)

  • Apr 28, 2009, 03:48 PM
    shortstuff001
    Exponential Equations
    The question is

    2^x=2^2x-6

    Can you explain how to solve this step by step?
  • Apr 28, 2009, 05:48 PM
    Perito




    Later. I started but I don't have time right now to finish -- and I'm a bit confused. I might need to leave this one for galactus.
    For what it's worth, X is between 1.58496 and 1.58497, but I worked that out numerically rather than in a closed form.
  • Apr 29, 2009, 12:29 AM
    Zazonker

    2^x = 2^2x -6
    Rearrange:
    2^2x - 2^x - 6 = 0
    Factor:
    (2^x + 2)(2^x -3) = 0
    Set each factor to zero

    First factor:
    2^x +2 = 0
    2^x = - 2 ----- not possible

    Second factor:
    2^x - 3 = 0
    2^x = 3
    take the ln of both sides

    x ln(2) = ln(3)
    x = ln(3)/ln(2)

    x= 1.0986/.6931
    x=1.58
  • Apr 29, 2009, 06:01 AM
    Perito
    Quote:

    Originally Posted by Zazonker View Post



    Rearrange:


    Factor:


    Set each factor to zero

    First factor:

    ----- not possible

    Second factor:



    take the ln of both sides








    Converted to LaTeX for clarity.

  • All times are GMT -7. The time now is 10:56 PM.