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More logarithm problems

  1. Nov 22, 2003 #1
    [tex]log_10x-2[/tex]=0 ....... that's log base 10
    (answer: 100)

    ln(x+5)=ln(x-1)-ln(x+1)
    (answer: no solution. but why?)

    [tex]log_4x-log_4(x-1)[/tex]=1/2
    (answer: 2)
     
    Last edited by a moderator: Nov 22, 2003
  2. jcsd
  3. Nov 22, 2003 #2
    [tex]
    \ln(x+5)=\ln(x-1)-\ln(x+1)=\ln\left(\frac{x-1}{x+1}\right)
    [/tex]

    So,

    [tex]
    x+5=\frac{x-1}{x+1}
    [/tex]

    Solve this for x, and you will see that both possible values generate a negative argument in the second logarithm, so there is no valid solution.
     
  4. Nov 22, 2003 #3
    #1:


    [tex]\log_{10} x - 2 = 0[/tex]


    [tex]10^{\log_{10}x} = 10^{2}[/tex]


    [tex]x = 10^{2} = 100[/tex]

    #3:


    [tex]\log_{4}x - \log_{4}(x-1) = 0.5[/tex]

    [tex]\log_{4}\frac{x}{(x-1)} = 0.5[/tex]

    [tex]\frac{x}{(x-1)} = 2[/tex]

    [tex]x = 2x - 2[/tex]

    [tex]x = 2 (x > 1)[/tex]
     
    Last edited: Nov 22, 2003
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