Results 1 to 7 of 7

Math Help - Log question

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    England
    Posts
    25

    Log question

    Please can someone help with this problem: solve 2^2^x^+^1 - 5 (2^x) + 2 = 0 Now we were told to let y=2 but I can't figure out where the y is coming from? Is it just as a function of x?

    Thanks in advance for your help!

    Kris
    Last edited by Krislton; March 9th 2013 at 11:42 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    602
    Thanks
    167

    Re: Log question

    You might have misread your book and it is telling you to make the substitution y=2x
    If not, do that substitution anyway and solve for y, after you have solved for y then finish by using it to find x

    Hint: 22x+1=2(2x)2



    ​Edit: Beat plato by 1 minute :P
    Last edited by Shakarri; March 9th 2013 at 11:02 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1

    Re: Log question

    Quote Originally Posted by Krislton View Post
    Please can someone help with this problem: solve 2^(2x+1) - 5 (2^x) + 2 = 0. Now we were told to let y=2 but I can't figure out where the y is coming from? Is it just as a function of x?
    If we let y=2^x that equation becomes 2y^2-5y+2=0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2012
    From
    England
    Posts
    25

    Re: Log question

    Quote Originally Posted by Shakarri View Post
    You might have misread your book and it is telling you to make the substitution y=2x
    If not, do that substitution anyway and solve for y


    Hint: 22x+1=2(2x)2
    Yeah I think your correct, but I'm still useless at logs! Its not from a text book, but from an old exam paper I need to get to grips with ASAP. Once re-arranged how you've done it, Should I then apply logs to both sides?

    Thanks for your help,

    Kris.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2012
    From
    England
    Posts
    25

    Re: Log question

    Quote Originally Posted by Plato View Post
    If we let y=2^x that equation becomes 2y^2-5y+2=0.
    Thanks, it kind of makes sense but if substituting 2^x = y, how can you sub that for 2x when that's not to the power unlike the other term.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701

    Re: Log question

    Hello, Krislton!

    Evidently you don't understand the hint or the suggestions.
    I'll give you a walk-through . . .


    \text{Solve: }\:2^{2x+1} - 5 (2^x) + 2 \:=\: 0

    We have: . 2^{2x+1} - 5(2^x) + 2 \;=\;0

    . . . . . . . . 2^{2x}\!\cdot\! 2^1 - 5(2^x) + 2 \;=\;0

    . . . . . . . . 2(2^x)^2 - 5(2^x) + 2 \;=\;0

    Let y = 2^x\!:\qquad 2y^2 - 5y + 2 \;=\;0

    Factor: . . . . (y-2)(2y-1) \;=\;0

    And we have: . y = 2,\;y = \tfrac{1}{2}


    Back-substitute: . \begin{Bmatrix}2^x \,=\,2 & \Rightarrow & 2^x \,=\,2^1 & \Rightarrow & x \,=\,1 \\ \\[-3mm]  2^x \,=\,\frac{1}{2} & \Rightarrow & 2^x \,=\,2^{\text{-}1} & \Rightarrow & x \,=\,\text{-}1 \end{Bmatrix}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2012
    From
    England
    Posts
    25

    Re: Log question

    Quote Originally Posted by Soroban View Post
    Hello, Krislton!

    Evidently you don't understand the hint or the suggestions.
    I'll give you a walk-through . . .



    We have: . 2^{2x+1} - 5(2^x) + 2 \;=\;0

    . . . . . . . . 2^{2x}\!\cdot\! 2^1 - 5(2^x) + 2 \;=\;0

    . . . . . . . . 2(2^x)^2 - 5(2^x) + 2 \;=\;0

    Let y = 2^x\!:\qquad 2y^2 - 5y + 2 \;=\;0

    Factor: . . . . (y-2)(2y-1) \;=\;0

    And we have: . y = 2,\;y = \tfrac{1}{2}


    Back-substitute: . \begin{Bmatrix}2^x \,=\,2 & \Rightarrow & 2^x \,=\,2^1 & \Rightarrow & x \,=\,1 \\ \\[-3mm]  2^x \,=\,\frac{1}{2} & \Rightarrow & 2^x \,=\,2^{\text{-}1} & \Rightarrow & x \,=\,\text{-}1 \end{Bmatrix}
    Perfect, thanks very much I understand now!
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum