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    Please help on how to solve these two questions

    a) Calculate the value of y for which 2 log₃y - log₃(y+4) = 2

    b) Calculate the values of z for which log₃z = 4log₃3
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    Use the log laws:

    log(a) + log(b) = log(ab)
    n*log(a) = log(a^n)

    and remember that logs are the opposite/inverse of powers.
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    (Original post by tommm)
    Use the log laws:

    log(a) + log(b) = log(ab)
    n*log(a) = log(a^n)

    and remember that logs are the opposite/inverse of powers.
    a)2 log₃y - log₃(y+4) = 2
    log₃(y^2/y+4)=2

    b) log₃z = 4log₃3
    log₃z = log₃3^4
    log₃z = log₃81
    log₃z = 4

    Is what I've done correct so far?
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    (Original post by enviusl)
    a)2 log₃y - log₃(y+4) = 2
    log₃(y^2/y+4)=2

    b) log₃z = 4log₃3
    log₃z = log₃3^4
    log₃z = log₃81
    log₃z = 4

    Is what I've done correct so far?
    Yep, that's all correct

    Now, you have something simpler to solve - what happens if you raise both sides to the power 3?
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    Could you not go straight from

    \log_3{z} = \log_3{81}

    to a final solution
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    (Original post by TenOfThem)
    Could you not go straight from

    log_3z = log_381

    to a final solution
    Next piece of training. \log looks nicer than log.
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    (Original post by TenOfThem)
    \log_3z = \log_381
    See!
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    (Original post by Mr M)
    Next piece of training. \log looks nicer than log.
    so it does
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    (Original post by enviusl)
    a)2 log₃y - log₃(y+4) = 2
    log₃(y^2/y+4)=2

    b) log₃z = 4log₃3
    log₃z = log₃3^4
    log₃z = log₃81
    log₃z = 4

    Is what I've done correct so far?
    Yes.

    Now you know that if

    \log_ab=c

    then

    a^c=b


    \log_3z = 4 \implies z=3^4 = 81

    but you could have already seen that after you did line 3 of your working.
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    (Original post by tommm)
    Yep, that's all correct

    Now, you have something simpler to solve - what happens if you raise both sides to the power 3?
    a)2 log₃y - log₃(y+4) = 2
    log₃(y^2/y+4)=2
    (y^2/y+4) = 3^2
    y^2 = 9 (y+4)
    y^2 = 9y + 36
    y^2 -9y -36 = 0
    (y + 3)(y -12)=0
    y=-3,12
    y=12

    b) log₃z = 4log₃3
    log₃z = log₃3^4
    log₃z = log₃81
    log₃z = 4
    z=81
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    all correct
 
 
 
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