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Thread: logarithm

  1. #16
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    well you are almost to the solution:

    convert the base to 10
    log10 (x^2 + 4x + 3) / log10 3 = 1

    log10 (x^2 + 4x + 3) = log10 3

    comparing left and right hand side:
    x^2 + 4x + 3 = 3

    solving for x you will get x = 0, x = -4

    then you substitute for each value of x into the main equation and carry out your reasoning why which value of x is acceptable and which is not acceptable
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  2. #17
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    okay.. now i have 3 more problems on the studyguide that i just don't understand.. i really am trying though..

    1... log x-3log 2=2... i know the answer is 800, and i have no idae where to start.. and the x and the 2 are regular, they are not subscripts.

    2... log(6x+4)=2... the farthest i got was to put log in front of the 2..

    3... ln(x-2) + ln(x-5) = ln(x-1).. i don't know where to start there, but i know it involves some e functions, i think.!

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  3. #18
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    For the first one, well transform -3log2 into $\displaystyle -\log 2^3$
    Then, as it is log, it's base 10. So $\displaystyle 2=2*1=2*\log(10)=\log(10^2)$

    Hence $\displaystyle \log(x)-3\log(2)=2 \longleftrightarrow \log(x)-\log(8)=\log(100)$

    Can you continue ? I've given you all the formulas you needed


    $\displaystyle log(6x+4)=2$

    $\displaystyle log(6x+4)=log(10^2) \longleftrightarrow 6x+4=\dots$


    ln(x-2) + ln(x-5) = ln(x-1)

    group the logarithms in the left hand side, then solve.
    If you want to go with exp, then :

    $\displaystyle exp(ln(x-2)+ln(x-5))=exp(ln(x-1))$

    $\displaystyle =exp(ln(x-2))*exp(ln(x-5))=exp(ln(x-1))$

    $\displaystyle (x-2)(x-5)=(x-1)$
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  4. #19
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    what is exp?
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  5. #20
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    e power

    The e you talked about, exponential
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  6. #21
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    i'm so lost on this last one... it's not even funny! i don't understand it, at all!
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  7. #22
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    Ok so do it the right way :

    ln(x-2)+ln(x-5)=ln[(x-2)(x-5)]
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  8. #23
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    okay so what do you do with the ln(x+1) thats on the right side of the equal sign
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  9. #24
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    If $\displaystyle ln(a)=ln(b)$ then a=b

    According to it, you can say that (x-2)(x-5)=... ?
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  10. #25
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    okay so after you work it out,

    you get x^2-4x-11... right?

    and then because you can't factor that, use the quadratic equation???
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  11. #26
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    Hm...

    I get (x-2)(x-5)=x-1
    This makes x-7x+10=x-1

    x-8x+11=0

    This can be factored, yep, use the quadratic thing :

    x-8x+16-5=0

    (x-4)=5
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  12. #27
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    okay well it says the answer is { 6.236 }
    (1.764 is NOT a solution
    we cannot take the log
    of a negative number)

    i don't understand that...
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  13. #28
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    Since you have ln(x-2), ln(x-5) and ln(x-1), you must have :

    x-2>0 <=> x>2
    x-5>0 <=> x>5
    x-1>0 <=> x>1

    So finally, x has to be superior to 5. Hence 6.236 works, but not 1.764
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  14. #29
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    but how can you get the 6 number??
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  15. #30
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    (x-4)=5

    <=> $\displaystyle x-4=\sqrt{5}$ or $\displaystyle x-4=-\sqrt{5}$

    <=> $\displaystyle x=4+\sqrt{5}$ or $\displaystyle x=4-\sqrt{5}$

    And here is the miracle : $\displaystyle \sqrt{5} \approx 2.23606797749979$
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