Results 1 to 7 of 7

Thread: A rather basic log question I believe

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    8

    A rather basic log question I believe

    "Solve for x:"

    2 X 3^2x+3 = 3 X 2^3x+4

    (X is multiply)

    feel silly for posting this, is a simple log question i think, but its got me stumped.

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Just to clarify, you want to solve $\displaystyle 2\cdot 3^{2x}+3=3\cdot 2^{3x}+4.$ Is that correct? If so, what steps have you taken?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    682
    Thanks
    1
    it could be also:
    $\displaystyle 2(3)^{2x+3} = 3(2)^{3x+4}$

    in which case $\displaystyle x = -1$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    8
    sorry yeah the above is the way it's laid out - how did you get -1 ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    682
    Thanks
    1

    Cool well several ways to solve this..

    well several ways to solve this..

    but you can divide both sides by $\displaystyle 2^1$ and $\displaystyle 3^1$ to elimate the coeficients,

    then
    $\displaystyle 3^{2x+2} = 2^{3x+3}
    \Rightarrow
    (2x+2)\ln{3} = (3x+3)\ln{2}
    \Rightarrow
    \frac{2x+2}{3x+3} = \frac{\ln{2}}{\ln{3}}
    $
    after calculation we are left with

    $\displaystyle x(.107) = -.107\Rightarrow x= -1$
    Last edited by bigwave; Jul 15th 2010 at 02:05 PM. Reason: added steps
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    8
    ahhh it's so simple, i'm so stupid. haha thanks, i just didn't start with the basic, "dividng by 2" part. oops.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie Zharif93's Avatar
    Joined
    Jul 2010
    From
    KL
    Posts
    11
    2(3)^2x+3 = 3(2)^3x+4

    log 2 log 3^(2x+3) = log 3 log 2^(3x+4)

    (2x+3) log 3 = log 3
    (3x+4) log 2 = log 2

    2x log 3 + 3 log = log 3

    2x log 3 = log 3 - 3 log 3

    x(2 log 3) = -2 log 3

    x = -2 log 3
    /2 log 3

    x = -1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 21st 2011, 08:21 PM
  2. Basic Question In PDE
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Oct 24th 2010, 01:44 PM
  3. basic log question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 6th 2010, 04:11 PM
  4. basic question, thank you
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 16th 2008, 06:59 PM
  5. Basic Question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Mar 2nd 2008, 10:57 AM

Search Tags


/mathhelpforum @mathhelpforum