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Kilochan

  • one year ago

log[3](x-2)=log[3] 27 - log[3](x-4) -5^(log[5]1) I have tried this out and right now all I am getting is log[3] (x-2)(x-4)/27= -5^(log[5] 1) Is this somewhat right so far? I'm not sure what to do with the "log[3] 27" the answer is supposed to be- 3+sqrt(10)

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  1. RadEn
    • one year ago
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    hint : -5^(log[5] 1 = -1 now, setting -1 be log[3] (1/3)

  2. ByteMe
    • one year ago
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    HINT(also): \(\large log_327=a \Leftrightarrow 3^a=27 \) what's a= ???

  3. Kilochan
    • one year ago
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    a=3? D:

  4. RadEn
    • one year ago
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    yes, that's right.. then, ?

  5. RadEn
    • one year ago
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    well, i want make easier for u from ur solution above : log[3] (x-2)(x-4)/27= -5^(log[5] 1 setting right side be -1 it can be : log[3] (x-2)(x-4)/27= -1 then -1 = [3] (1/3), therefore it can be log[3] (x-2)(x-4)/27= [3] (1/3) cancel out the log[3] to both sides (x-2)(x-4)/27= 1/3 or (x-2)(x-4) = 27 * 1/3 (x-2)(x-4) = 9 now, solve for x

  6. RadEn
    • one year ago
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    * log[3] (x-2)(x-4)/27= log [3] (1/3)

  7. Kilochan
    • one year ago
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    Ok! I was close. I for for to put log[3] on both sides. But now that I see where I went wrong, I will finish solving for x and let you know if I have any problems. Thank you so much for your help!

  8. Kilochan
    • one year ago
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    *forgot to put log[3]

  9. RadEn
    • one year ago
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    okay... to get solution of (x-2)(x-4) = 9 simplify, frist. then use quadratic formuka : x = {-b +- sqrt(b^2-4ac)}/2a if u got 2 solution's, that one positive and the other negative, just take that positive value, because for x negative undefined for log[3] (x-2)

  10. Kilochan
    • one year ago
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    I finished the problem! I got the correct answer! Thanks for your help! One last question. for -5^(log[5] 1, did you plug that in a calculator? I tried that and I'm getting the number 3.

  11. RadEn
    • one year ago
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    no, i have used the property of logarithm : |dw:1361529157735:dw|

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