Results 1 to 4 of 4

Math Help - How should I solve these equations?

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    3

    How should I solve these equations?

    Hey all,
    I am trying to solve these 2 equations but I can't.
    27^x + 12^x = 2 * 8^3
    and,
    3 * 16 ^x + 37 * 36^x = 26 * 81^x
    thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by mehdi View Post
    Hey all,
    I am trying to solve these 2 equations but I can't.
    27^x + 12^x = 2 * 8^3
    and,
    3 * 16 ^x + 37 * 36^x = 26 * 81^x
    thanks in advance.
    I can show you how to do the 2nd one - but it is kind of tricky:

    3 * 16 ^x + 37 * 36^x = 26 * 81^x

     37 \cdot 6^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}

     37 \cdot 2^{2x} \cdot 3^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x} divide by 2^{2x}

     37 \cdot  3^{2x}+3 \cdot 2^{2x} = 26 \cdot 3^{4x} \cdot \dfrac1{2^{2x}} divide by 3^{2x}

     37 +3 \cdot \left(\dfrac23\right)^{2x} = 26 \cdot 3^{2x} \cdot \dfrac1{2^{2x}} = 26 \cdot \left(\dfrac32\right)^{2x}

    Now substitute y = \left(\dfrac32\right)^{2x}. The equation becomes:

    37+3 \cdot \dfrac1y=26 \cdot y multiply by y:

    26y^2-37y-3=0

    yields y = \frac32~\vee~y=-\frac1{13}

    Since y is a power with a positive base the 2nd solution can't be correct.

    Now re-substitute:

    \left(\dfrac32\right)^{2x} = \dfrac32

    That means 2x = 1~\implies~\boxed{x = \dfrac12}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    ...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, mehdi!

    Another approach to the second problem . . .


    3\cdot16^x + 37\cdot36^x \:=\: 26\cdot81^x

    We have: . 3(2^4)^x + 37(2^2\!\cdot\!3^3)^x - 26(3^4)^x \;=\;0

    . . . . . . . 3(2^{4x}) + 37(2^{2x})(3^{2x}) - 26(3^{4x}) \;=\;0

    Factor: . \bigg[2^{2x} + 13\!\cdot\!3^{2x}\bigg]\,\bigg[3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x}\bigg] \;=\;0



    We have two equations to solve:


    2^{2x} + 13\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 2^{2x} \:=\:\text{-}13\!\cdot\!3^{2x}

    . . \dfrac{2^{2x}}{3^{2x}} \:=\:\text{-}13 \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} \:=\:\text{-}13\;\hdots\;\text{ no real roots}


    3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 3\!\cdot\!2^{2x} \:=\:2\!\cdot\!3^{2x}

    . . \dfrac{2^{2x}}{3^{2x}} \:=\:\dfrac{2}{3} \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} =\:\left(\dfrac{2}{3}\right)^1 \quad\Rightarrow\quad 2x \:=\:1

    . . \text{Therefore: }\:\boxed{x \:=\:\frac{1}{2}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. equations solve
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 30th 2011, 06:46 AM
  2. Equations - Can't solve it!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 22nd 2009, 02:22 AM
  3. help me to solve the equations
    Posted in the Algebra Forum
    Replies: 10
    Last Post: October 23rd 2008, 10:26 PM
  4. Solve for X in equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 4th 2007, 02:51 AM
  5. solve equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 24th 2006, 09:40 AM

Search Tags


/mathhelpforum @mathhelpforum