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    hey there

    I have a question and theres one part of it that just is not making any sense

    "Find the exact values of x and y given that logyx=3 and Log3x-log3y=5"

    okay so I did:
    logyx=3
    y3=x

    log3x-log3y=5
    log3x/y=5
    Log3y2=5
    y2=35
    y=35/2 <---why can we take the 2 from the left and divide it by the five? I didn't know we could do this

    when I sub y=35/2 in y^3=x i get 3788 and not the answer of 3^(15/2) which it says on the mark scheme

    any ideas anyone? thank you
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    (Original post by upthegunners)
    hey there

    I have a question and theres one part of it that just is not making any sense

    "Find the exact values of x and y given that logyx=3 and Log3x-log3y=5"

    okay so I did:
    logyx=3
    y3=x

    log3x-log3y=5
    log3x/y=5
    Log3y2=5
    y2=35
    y=35/2 <---why can we take the 2 from the left and divide it by the five? I didn't know we could do this

    when I sub y=35/2 in y^3=x i get 3788 and not the answer of 3^(15/2) which it says on the mark scheme

    any ideas anyone? thank you
    ....

    a^2 = x

    a = x^(1/2)
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    (Original post by upthegunners)
    hey there

    I have a question and theres one part of it that just is not making any sense

    "Find the exact values of x and y given that logyx=3 and Log3x-log3y=5"

    okay so I did:
    logyx=3
    y3=x

    log3x-log3y=5
    log3x/y=5
    Log3y2=5
    y2=35
    y=35/2 <---why can we take the 2 from the left and divide it by the five? I didn't know we could do this

    when I sub y=35/2 in y^3=x i get 3788 and not the answer of 3^(15/2) which it says on the mark scheme

    any ideas anyone? thank you
    y=35/2 in y^3=x << this equals y=315/2

    simple 5 x 3

    oh and if you put 3^7.5 in your calculator you get 3788

    ryan
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    (Original post by ryanb97)
    y=35/2 in y^3=x << this equals y=315/2

    simple 5 x 3

    ryan
    is the rule of indices not that (5/2)3 = 125/8 ?
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    (Original post by 2710)
    ....

    a^2 = x

    a = x^(1/2)
    how does this work? I dont remember ever learning this indices rule
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    you have (keep it in this form): y^{2}=3^{5}=243=81 \times 3

    so, take the square root of 81 x 3 = \pm 9\sqrt{3}

    sub that (each) into x=y^3
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    (Original post by upthegunners)
    is the rule of indices not that (5/2)3 = 125/8 ?
    That is one rule of indices

    Another (more relevant here) is that (a^b)^c = a^{bc}
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    (Original post by upthegunners)
    how does this work? I dont remember ever learning this indices rule
    The rule that squaring is the opposite of square rooting

    I think you probably have learnt that
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    (Original post by TenOfThem)
    That is one rule of indices

    Another (more relevant here) is that (a^b)^c = a^{bc}
    thanks so much!

    So what is the rule that would allow me to cube it?
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    (Original post by upthegunners)
    how does this work? I dont remember ever learning this indices rule
    It's the rule of square roots,
    a^2=b then a=\sqrt b=b^{\frac{1}{2}}
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    (Original post by Robbie242)
    It's the rule of square roots,
    a^2=b then a=\sqrt b=b^{\frac{1}{2}}
    I always thought that if a fraction was raised to the power 3 you cubed both the numerator and the denominator
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    (Original post by upthegunners)
    I always thought that if a fraction was raised to the power 3 you cubed both the numerator and the denominator
    Depends what fraction if the numerator>1 or <-1 then it will have an effect, if it is =1 or -1 then it will have an effect as (-1)^3=-1. In short I just just cube it so (\dfrac{a}{b})^3=\dfrac{a^{3}}{b  ^{3}}=\frac{a}{b}\times\frac{a}{  b}\times\frac{a}{b}
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    (Original post by Robbie242)
    Depends what fraction
    how can it be different?

    A fraction is a fraction
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    (Original post by upthegunners)
    hey there

    I have a question and theres one part of it that just is not making any sense

    "Find the exact values of x and y given that logyx=3 and Log3x-log3y=5"

    okay so I did:
    logyx=3
    y3=x

    log3x-log3y=5
    log3x/y=5
    Log3y2=5
    y2=35
    y=35/2 <---why can we take the 2 from the left and divide it by the five? I didn't know we could do this

    when I sub y=35/2 in y^3=x i get 3788 and not the answer of 3^(15/2) which it says on the mark scheme

    any ideas anyone? thank you
    From the identities of
    \displaystyle \left (a^n\right )^k=\left (a^k\right )^n=a^{n\cdot k}
    and
    \displaystyle \left (a^{p}\right )^{\frac{1}{q}}=\sqrt[q]{a^p}=\left (\sqrt [q]{a}\right )^p=\left (a^{\frac{1}{q}}\right )^p=a^{\frac{1}{q}\cdot p}=a^{\frac{p}{q}}
    SO
    y^2=3^5
    taking squre root
    \displaystyle \left (y^2\right )^{\frac{1}{2}}=\left (3^5\right )^{\frac{1}{2}}
    \displaystyle y^{2\cdot \frac{1}{2}}=y^1=y=3^{\frac{5}{2  }}
    Similarly
    \displaystyle x=y^3=\left (3^{\frac{5}{2}}\right )^3=3^{\frac{5}{2}\cdot 3}=3^{\frac{15}{2}}
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    (Original post by upthegunners)
    how can it be different?

    A fraction is a fraction
    yes I mean if it were 1 then cubing it would not alter it at all, but yes your right a fraction is indeed a fraction
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    (Original post by upthegunners)
    I always thought that if a fraction was raised to the power 3 you cubed both the numerator and the denominator
    You do...
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    (Original post by 2710)
    You do...
    then why not in the original case?
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    (Original post by upthegunners)
    then why not in the original case?
    Which original case? I dont see any fraction?

    y^2=3^5

    y = 3^(5/2)

    Wheres the fraction u are talking about?
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    (Original post by upthegunners)
    then why not in the original case?
    Because you are not raising a fraction to a power
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    Oh i see. Because as above has said, you are not raising the fraction to a power if u look carefully
 
 
 
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