Results 1 to 4 of 4

Thread: Solve for x...

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    145

    Solve for x...

    Hi,

    Solve for x:

    $\displaystyle (a.)\,4^x - 13(2^x) + 40 = 0$

    This is supposed to have two solutions (quadratic), but I've only progressed two steps. I believe these two steps to be incorrect. Argh!

    $\displaystyle 4^x - 26^x + 40 = 0$
    $\displaystyle -22^x + 40 = 0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    They are incorrect.

    $\displaystyle 13(2^x)\neq 26^x$ and $\displaystyle 4^x - 26^x \neq -22^x$

    Recall BODMAS and 'adding like terms' for reasons why.

    Try this

    $\displaystyle \displaystyle 4^x - 13(2^x) + 40 = 0$

    $\displaystyle \displaystyle (2^2)^x - 13(2^x) + 40 = 0$

    $\displaystyle \displaystyle (2)^{2x} - 13(2^x) + 40 = 0$

    $\displaystyle \displaystyle (2^x)^{2} - 13(2^x) + 40 = 0$

    Now making $\displaystyle 2^x = a$ you get

    $\displaystyle \displaystyle a^{2} - 13a + 40 = 0$

    Can you continue? What do you get?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    145
    Quote Originally Posted by pickslides View Post
    Can you continue? What do you get?
    Thanks.

    $\displaystyle (a - 5)(a - 8) = 0$
    $\displaystyle a = 5; a = 8$

    $\displaystyle 2^x\,=\,5; 2^x\,=\,8$

    $\displaystyle \log(2^x) = \log\,5;\,\,\,\,\,\,\,\,\,\log\,(2^x) = \log\,8$

    $\displaystyle x = \frac{\log\,5}{\log\,2};\,\,\,\,\,x = \frac{\log\,8}{\log\,2}$

    $\displaystyle x \approx 2.32;\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3$

    $\displaystyle 13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x$
    'Like' here means: of the same base?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by Hellbent View Post

    $\displaystyle 13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x$
    'Like' here means: of the same base?
    Yep, like this

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

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Jul 16th 2010, 10:29 PM
  2. Replies: 1
    Last Post: Jun 9th 2009, 10:37 PM
  3. how do I solve this?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jan 22nd 2009, 06:21 PM
  4. How could i solve?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 2nd 2009, 02:18 PM
  5. Solve for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jan 1st 2009, 12:33 PM

Search Tags


/mathhelpforum @mathhelpforum