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    Q6 ii.

    Logx^10-2Log(x^3/4)=4Log(2x) (show that this is the case)

    Could someone please help me?

    how did i get negged for this? -.-
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    (Original post by Davelittle)
    Q6 ii.

    Logx^10-2Log(x^3/4)=4Log(2x) (show that this is the case)

    Could someone please help me?
    Remember that:
    \log_a(x^n) = n\log_a(x)
    \log_a(\frac{x}{y}) = \log_a(x) - log_a(y)
    and \log_a(xy) = \log_a(x) + log_a(y)
    Be careful with the signs when you split up the logarithms
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    (Original post by joostan)
    Remember that:
    \log_a(x^n) = n\log_a(x)
    \log_a(\frac{x}{y}) = \log_a(x) - log_a(y)
    and \log_a(xy) = \log_a(x) + log_a(y)
    Be careful with the signs when you split up the logarithms
    i tried that and ended up with 4LogX-Log16 :/

    please could you try it and see if you can do it (my exam is tomorrow so i am really worried)
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    (Original post by Davelittle)
    i tried that and ended up with 4LogX-Log16 :/

    please could you try it and see if you can do it (my exam is tomorrow so i am really worried)
    2log(x^3/4) = log(x^6/16)

    so

    log(x^10) - 2log(x^3/4) = log(x^10/(x^6/16)) = log(16x^4)

    Can you finish it from here?
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    (Original post by davros)
    2log(x^3/4) = log(x^6/16)

    so

    log(x^10) - 2log(x^3/4) = log(x^10/(x^6/16)) = log(16x^4)

    Can you finish it from here?
    Log(2x)^4

    4Log(2x)

    Thanks so much

    p.s. i want to rep you but i already repped you recently
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    (Original post by Davelittle)
    Log(2x)^4

    4Log(2x)

    Thanks so much

    p.s. i want to rep you but i already repped you recently
    no problem!
 
 
 
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