Results 1 to 11 of 11

Math Help - I have 4^(x+1) .. how to proceed

  1. #1
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    I have 4^(x+1) .. how to proceed

    Hi!


    4^(x+1) + 3 * 4^x = 112

    Should I use log or something?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Twig View Post
    Hi!


    4^(x+1) + 3 * 4^x = 112

    Should I use log or something?
    logarithms? no. but good thinking. you should be thinking logarithms for things like this, but in this case, logarithms are not necessary

    note that 4^{x + 1} = 4 \cdot 4^x

    can you continue now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    hi

    I did solve it, but I doubt I did it "correctly".

    I got 4 * 4^x + 3 * 4^x = 112

    I divided all terms with 4^x

    so I have 7 = 112/4^x

    this gives me 4^x = 112/7

    now I used logarithms to solve for x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Twig View Post
    I did solve it, but I doubt I did it "correctly".

    I got 4 * 4^x + 3 * 4^x = 112

    I divided all terms with 4^x

    so I have 7 = 112/4^x


    this gives me 4^x = 112/7

    now I used logarithms to solve for x
    what is in red is completely unnecessary

    now simplify first. 112/7 = 16

    so you have 4^x = 16

    do you really need logarithms for that?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    hi

    so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ?

    and just from there take 112/7 = 16

    and 4^x = 16
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Twig View Post
    hi

    so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ?

    and just from there take 112/7 = 16

    and 4^x = 16
    yes (you divide both sides by 7)

    now what is the solution?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    x = 2
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Twig View Post
    x = 2
    .....you didn't use logarithms, did you?

    are we good now? do we understand the concept?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    thanks a lot
    Yes I did.

    I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Twig View Post
    thanks a lot
    Yes I did.

    I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x
    yes. you can essentially treat the 4^x as a variable. just as how if you had 4x + 3x you would get 7x. the idea is the same

    or if you don't want to think of it that way, you can think of it as factoring out the 4^x
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    alright cool =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Polynomial Problem and not a clue how to proceed
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: March 11th 2011, 08:50 AM
  2. Replies: 2
    Last Post: April 23rd 2009, 11:41 AM
  3. Replies: 3
    Last Post: August 30th 2008, 06:17 AM

Search Tags


/mathhelpforum @mathhelpforum