Results 1 to 7 of 7

Math Help - Solving a (Tough) Exponential Function

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    126

    Solving a (Tough) Exponential Function

    So, please don't ignore me cause I'm only going to give you the equation but I've tried several different routes of attacking this equation and I can't seem to figure it out!

    4^x + 6(4^(-x)) = 5
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello,
    Quote Originally Posted by mike_302 View Post
    So, please don't ignore me cause I'm only going to give you the equation but I've tried several different routes of attacking this equation and I can't seem to figure it out!

    4^x + 6(4^(-x)) = 5
    Multiplying both sides by 4^x we get 4^{2x}+6=5\times 4^x that is to say \left(4^x\right)^2-5\times4^x+6=0. Can you solve this equation ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    126
    Yeeeeeaaaa, I can see how that works.... thanks a bunch! I was aiming for that quadratic form a couple of times, but I kept seeing the whole 4^2x as... simply that, 4^2x ... I didn't think of factoring it out to get "squared"

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, Mike!

    Solve for x\!:\;\;4^x + 6\cdot4^{-x} \:=\: 5

    Multiply by 4^x\!:\;\;4^{2x} + 6 \:=\:5\cdot4^x\quad\Rightarrow\quad 4^{2x} - 5\cdot 4^x + 6 \:=\:0

    Factor: . \left(4^x-2\right)\left(4^x - 3\right) \:=\:0


    Then we have:

    . . 4^x - 2 \:=\:0 \quad\Rightarrow\quad 4^x \:=\:2 \quad\Rightarrow\quad x \:=\:\log_4(2) \quad\Rightarrow\quad\boxed{ x  \:=\:\frac{1}{2}}

    . . 4^x - 3 \:=\:0\quad\Rightarrow\quad 4^x \:=\:3 \quad\Rightarrow\quad x \:=\:\log_4(3)\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\ln(3)}{\ln(4)}}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2008
    Posts
    126
    Thanks both of you! I shall remember this little..... trickery of 2x and x^2 (as exponents)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2008
    Posts
    126
    Let me know if I'm making a mistae by just reposting in this thread.... It's the exact same question, so I didn't think it was worthy of a thread wit hthe same name.

    Here's the question:

    log(3x-5)base4 = log(11)base4 + log(2)base4

    I got it to

    log(3x-5)base4 = log(22)base4
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Feb 2008
    Posts
    126
    AHH! I figured it out.... Same bases. Nevermind :P

    22=3x-5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Exponential and Log. Function
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 21st 2010, 04:18 AM
  2. tough inverse function problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 30th 2009, 03:02 PM
  3. Tough Trigonometric Function Question!!
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 18th 2009, 03:15 AM
  4. Replies: 0
    Last Post: February 8th 2008, 09:28 AM
  5. Solving exponential
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 27th 2007, 03:27 PM

Search Tags


/mathhelpforum @mathhelpforum