Results 1 to 10 of 10

Math Help - few questions

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    9
    Awards
    1

    few questions

    logarithms
    (in each case check your answer by subtituting back the value for x)

     6.5^x =8.4^{(4x-10)}


    differentials

     s=4e^{3t}  -e^{-2.5t} w.r.t. t

    y=4sin2\theta-3cos4\theta  w.r.t. \theta

    integrate

    \int(2x^3-3x^2+5x-2)dx

    thanks
    Last edited by Problem_Solver11; June 28th 2008 at 03:24 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,862
    Thanks
    742
    Hello, Problem_Solver11!

    I see that you're still learning LaTeX . . .


     6.5^x \:=\:8.4^{4x-10}

    Take logs: . \ln\left(6.5^x\right) \;=\;\ln\left(8.4^{4x-10}\right) \quad\Rightarrow\quad x\ln(6.5) \;=\;(4x-10)\ln(8.4)

    . . . . x\ln(6.5) \;=\;4x\ln(8.4) - 10\ln(8.4) \quad\Rightarrow\quad 4x\ln(8.4) - x\ln(6.5) \;=\;10\ln(8.4)

    Factor: . x\bigg[4\ln(8.4) - \ln(6.5)\bigg] \;=\;10\ln(8.4) \quad\Rightarrow\quad x \;=\;\frac{10\ln(8.4)}{4\ln(8.4) - \ln(6.5)}


    Therefore: . x \;=\;3.20462545\hdots

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    9
    Awards
    1
    yeh haha I am I only just learnt it well still are, once I keep doing them im sure I'll be ok

    and thanks ^

    EDIT: in differentials the s=4e3t the t is ment to be low case like the 3 how do I do this :X
    Second EDIT: just done the LaTeX right now
    Last edited by Problem_Solver11; June 28th 2008 at 03:25 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Problem_Solver11 View Post
    logarithms
    (in each case check your answer by subtituting back the value for x)

     6.5^x =8.4^{(4x-10)}


    differentials

     s=4e^{3t}  -e^{-2.5t} w.r.t. t

    y=4sin2\theta-3cos4\theta  w.r.t. \theta

    integrate

    \int(2x^3-3x^2+5x-2)dx

    thanks
     s=4e^{3t}  -e^{-2.5t}
    \frac{\mathrm{d}(e^{ax})}{\mathrm{d}x} = ae^{ax} Where a is a constant.
    \frac{\mathrm{d}s}{\mathrm{d}t} = 12e^{3t} + 2.5e^{-2.5t}


    y=4sin2\theta-3cos4\theta
    For Such, \frac{\mathrm{d}(\sin Ax)}{\mathrm{d}\theta} = A\cos Ax, \frac{\mathrm{d}(\cos Ax)}{\mathrm{d}\theta} = - A\sin Ax
    \frac{\mathrm{d}y}{\mathrm{d}\theta} = 8\cos 2\theta + 12 \sin 4 \theta


    \int(2x^3-3x^2+5x-2)dx
    For Such, \int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c
    = \frac{2x^4}{4} + \frac{3x^3}{3} + \frac{5x^2}{2} - \frac{2x}{1} + c
    = \frac{x^4}{2} + 3x^3 + \frac{5x^2}{2} - 2x + c
    Last edited by Simplicity; June 29th 2008 at 04:26 AM. Reason: Correction
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2008
    Posts
    9
    Awards
    1
    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi

    Quote Originally Posted by Air View Post
    [..]
    For Such, { \color{red}\int}t^a{\color{red}\,\mathrm{d}t} = \frac{x^{a+1}}{a+1}{\color{red}+C}
    [...]
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2008
    Posts
    9
    Awards
    1
    Quote Originally Posted by flyingsquirrel View Post
    Hi
    why did u quote her work? was there a mistake?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by Problem_Solver11 View Post
    why did u quote her work? was there a mistake?
    Yes, what's in red in the quotation was missing... but you can't see it now since Air has edited her post to correct it.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Problem_Solver11 View Post
    why did u quote her work? was there a mistake?
    When writing the general form for integral, I wrote:

    x^a = \frac{x^{a+1}}{a+1}

    and it would have made proper sense to write:

    \int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c

    Hence I changed it.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jun 2008
    Posts
    9
    Awards
    1
    Quote Originally Posted by Air View Post
    When writing the general form for integral, I wrote:

    x^a = \frac{x^{a+1}}{a+1}

    and it would have made proper sense to write:

    \int x^a \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + c

    Hence I changed it.
    oh ok I see
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 05:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 05:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 04:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 08:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 02:53 AM

Search Tags


/mathhelpforum @mathhelpforum