Results 1 to 4 of 4

Math Help - anti-logging

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    93

    anti-logging

    Hi:
    If I have an equation of the sort

    <br /> <br />
\ln \left( {3y - 1} \right) = \frac{x}{2}<br />

    and I want to solve for y, I can anti-log which gives:

    <br />
3y - 1 = e^{\frac{x}{2}} <br />

    The problem is, I can't understand how the x/2 has now become a power. Because if I multiply ln by e to give lne=1, and I do the same for the RHS which gives ex/2, not <br />
e^{\frac{x}{2}} <br />

    can someone help me to understand this please
    thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2010
    Posts
    93
    ahhh!
    I may be able to answer my own question here.
    Is it because

    <br />
\ln x = y{\rm{  }} \Leftrightarrow {\rm{  e}}^y  = x<br />
?

    So I shouldn't be thinking about algebraically multiplying both sides by e, but instead be applying the log laws?
    Last edited by stealthmaths; May 3rd 2010 at 06:43 AM. Reason: forgotten statement
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by stealthmaths View Post
    Hi:
    If I have an equation of the sort

    <br /> <br />
\ln \left( {3y - 1} \right) = \frac{x}{2}<br />

    and I want to solve for y, I can anti-log which gives:

    <br />
3y - 1 = e^{\frac{x}{2}} <br />

    The problem is, I can't understand how the x/2 has now become a power. Because if I multiply ln by e to give lne=1, and I do the same for the RHS which gives ex/2, not <br />
e^{\frac{x}{2}} <br />
    There is no multiplication involved, just the definition of a logarithm.

    In other language \ln(x) is not a number like thing by which x is multiplied but an operator which transforms its input ( x) in to its output as specified by the definition of the logarithm.

    CB

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2010
    Posts
    93
    ahhh. I see.
    That's very clear. Thank you Captain.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. anti-derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 4th 2010, 11:24 PM
  2. anti-derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 4th 2010, 05:15 PM
  3. Anti-differentiation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 3rd 2008, 09:55 AM
  4. Anti derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 22nd 2008, 10:10 AM

Search Tags


/mathhelpforum @mathhelpforum