Results 1 to 11 of 11

Math Help - Exponential problem

  1. #1
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92

    Exponential problem

    Guys, Just a quick one:

    Can anyone help me with e^((x+2lnx)) which is part of a bigger problem I'm working on?

    I think that it is e^x + x^2 but apparently this is wrong. I can't see how to move on as I though if you took the anti-log this is what you'd get.

    Your help is much needed.

    D
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by dojo View Post
    Guys, Just a quick one:

    Can anyone help me with e^((x+2lnx)) which is part of a bigger problem I'm working on?

    I think that it is e^x * x^2 but apparently this is wrong.<<<<< correct! See my correction
    I can't see how to move on as I though if you took the anti-log this is what you'd get.

    Your help is much needed.

    D
    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    Does that make it xe^2x then?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by dojo View Post
    Does that make it xe^2x then?
    e^{x+2ln(x)} = e^x \cdot e^{2\ln(x)} = e^x \cdot x^2

    You can't simplify this term any more.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    Thanks! This is part of another question which I stuck on.

    ln (3+2y) - ln (3-2y) = (x + 2lnx)

    Express y in terms of x in the form y = f(x)

    I can remove the logs and end up with :

    (3+2y)/(3-2y) = e^x * x^2

    But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

    Where should I go from here?

    Thanks D
    Follow Math Help Forum on Facebook and Google+

  6. #6
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by dojo View Post
    Thanks! This is part of another question which I stuck on.

    ln (3+2y) - ln (3-2y) = (x + 2lnx)

    Express y in terms of x in the form y = f(x)

    I can remove the logs and end up with :

    (3+2y)/(3-2y) = e^x * x^2

    But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

    Where should I go from here?

    Thanks D
    e^x and x^2 can be treated as a normal fraction. Multiply both sides by (3-2y) and expand before isolating y
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by dojo View Post
    Thanks! This is part of another question which I stuck on.

    ln (3+2y) - ln (3-2y) = (x + 2lnx)

    Express y in terms of x in the form y = f(x)

    I can remove the logs and end up with :

    (3+2y)/(3-2y) = e^x * x^2

    But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

    Where should I go from here?

    Thanks D
    \dfrac{3+2y}{3-2y} = e^x \cdot x^2

    Multiply by the denominator, seperate the terms: All terms containing y at the RHS, all other at the LHS:

    3+2y = (e^x \cdot x^2)(3-2y)~\implies~3-3(e^x \cdot x^2)=-2y-2y(e^x \cdot x^2)

    Now factor out y at the RHS:

    3-3(e^x \cdot x^2)=y(-2-2(e^x \cdot x^2))

    Divide by the bracket of the RHS to get y. Simplify this term a little bit more.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    OK. I think I might have it, could you check through these calculations to see if I'm right? Excuse if this looks a it messy...

    ln (3+2y) - ln (3-2y) = (x + 2lnx)

    ln ((3+2y)/(3-2y) = (x + 2lnx)

    (3+2y)/(3-2y) = e^x * x^2

    3+2y = (e^x * x^2) (3 - 2y)

    3+2y = 9ex^2x - 6e^xyx^2 - 6e^xyx^2 + 4e^xy^2x^2

    Simplified...

    3+2y = ex^2x (9 - 8y^2)

    (3+2y)/(9 - 8y^2) = ex^2x

    (3-2)/y = -72ex^2x

    1/(-72ex^2x) = y
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    Sorry, Ignore that last post. Your help in the reply before didn't open on my Mac. I can't see how you would tidy up the LHS though...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by dojo View Post
    OK. I think I might have it, could you check through these calculations to see if I'm right? Excuse if this looks a it messy...

    ln (3+2y) - ln (3-2y) = (x + 2lnx)

    ln ((3+2y)/(3-2y) = (x + 2lnx)

    (3+2y)/(3-2y) = e^x * x^2

    3+2y = (e^x * x^2) (3 - 2y) <<<<<< OK

    3+2y = 9ex^2x - 6e^xyx^2 - 6e^xyx^2 + 4e^xy^2x^2 <<<<<<<< Sorry, but NO!

    Simplified...

    ...
    I'll take your last correct line:

    3+2y = (e^x * x^2) (3 - 2y)

    3+2y = 3(e^x * x^2) - 2y(e^x * x^2)

    3 - 3(e^x * x^2) = -2y- 2y(e^x * x^2)

    3 - 3(e^x * x^2) = y(-2- 2(e^x * x^2))

    \dfrac{3 - 3(e^x * x^2)}{-2-  2(e^x * x^2)} = y

    -\dfrac32 \cdot \dfrac{1 - (e^x * x^2)}{1+(e^x * x^2)} = y

    \dfrac32 \cdot \dfrac{(e^x * x^2)+1}{(e^x * x^2)-1} = y
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    I see my error now; I though it was just an addition expansion. I get this and now and am happy with the technique. Thanks for your advice.

    Regards
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 9th 2011, 09:54 AM
  2. Exponential problem
    Posted in the Algebra Forum
    Replies: 7
    Last Post: June 22nd 2010, 08:04 AM
  3. Help: 2 exponential problem
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 19th 2009, 01:41 PM
  4. Exponential Problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 24th 2009, 05:11 PM
  5. Exponential problem
    Posted in the Algebra Forum
    Replies: 0
    Last Post: October 12th 2008, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum