Results 1 to 3 of 3

Math Help - Logarithm help required

  1. #1
    Junior Member
    Joined
    Feb 2009
    From
    Fiery pits of confusion
    Posts
    36

    Logarithm help required

    Use natural lograithms to solve each equation. Round answers to the nearest hundreth.
    4e^x=10
    4^x=ln10
    I am stuck here. Could anybody usher me along?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, puzzledwithpolynomials!

    Use natural lograithms to solve each equation. Round answers to the nearest hundreth.

    (1)\;\;4e^x\:=\:10
    Divide by 4: . e^x \:=\:2.5

    \text{Take logs: }\;\ln(e^x) \:=\:\ln(2.5) \quad\Rightarrow\quad x\!\cdot\!\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(2.5)

    Therefore: . x \:=\:\ln(2.5) \:\approx\:0.92



    (2)\;\;4^x\:=\:\ln10

    Take logs: . \ln(4^x) \:=\:\ln(\ln10) \quad\Rightarrow\quad x\!\cdot\!\ln4 \:=\:\ln(\ln10)

    Therefore: . x \:=\:\frac{\ln(\ln10)}{\ln4} \:\approx\:0.60

    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by puzzledwithpolynomials View Post
    Use natural lograithms to solve each equation. Round answers to the nearest hundreth.
    4e^x=10
    4^x=ln10
    I am stuck here. Could anybody usher me along?
    1. Divide through by 4 and then take the logarithm to base e:

    e^x = \frac{10}{4} = \frac{5}{2}

    x = ln{(\frac{5}{2})}

    2. Change of base rule: log_a(b) = \frac{log_c(b)}{log_c(a)}

    Take logs again, this time I will pick base 4 to demonstrate the change of base rule:

    x\log_4(4) = log_4{(ln(10))}

    x = log_4{(ln(10))}

    However if your calculator does not give you option to use base 4 then use the change of base rule above. In this case a = 4, b = ln(10) and c = 10

    log_4{(ln(10))} = \frac{log_{10}{(ln(10))}}{log_{10}(4)}


    edit: using base e or 10 instead of base 4 would be easier, I just wanted to demonstrate the change of base rule
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 18th 2011, 07:04 AM
  2. Some help required please
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 9th 2009, 01:02 AM
  3. Help required
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 24th 2009, 09:16 AM
  4. Help Required !!
    Posted in the Calculus Forum
    Replies: 6
    Last Post: January 5th 2008, 12:02 PM
  5. logarithm help required...
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 28th 2006, 01:39 PM

Search Tags


/mathhelpforum @mathhelpforum