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New Maths GCSE 9 -1 Exam Style Question, NEED HELP!!! Watch

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    Hey Everyone,
    I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
    3^2x = 1/81

    Any help?
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    (Original post by NothingButWaleed)
    Hey Everyone,
    I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
    3^2x = 1/81

    Any help?
    At GCSE, you can't rearrange to solve these.

    Start with 3^x = 81. Can you see what x must be here?

    Then what about 3^x = \frac{1}{81}. Using your answer above, what do you think x must be?

    Finally can you use that to answer your question?

    Please post all your thoughts / working if you get stuck.
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    (Original post by notnek)
    At GCSE, you can't rearrange to solve these.

    Start with 3^x = 81. Can you see what x must be here?

    Then what about 3^x = \frac{1}{81}. Using your answer above, what do you think x must be?

    Finally can you use that to answer your question?

    Please post all your thoughts / working if you get stuck.
    OK i understand but then it says 2x?
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    Look up logarithms
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    (Original post by NothingButWaleed)
    OK i understand but then it says 2x?
    E.g. if you had 2^{2x}=16

    Since the solution to 2^x = 16 is 4, for the above equation you need 2x=4 so x=2.

    Can you try that for your question?
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    (Original post by Danielelliston)
    Look up logarithms
    They're not part of GCSE. This question is meant to be done by considering powers of 3.
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    (Original post by NothingButWaleed)
    Hey Everyone,
    I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
    3^2x = 1/81

    Any help?
    re write the rhs with a base of 3 to start with
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    OHHHHHH!!!!,

    So if there is a 3, i need to do 3^4 to get 81, but if there is a 2x i need an integer which will give me 4 when multiplied by 2, that number is 2
    But to get a fraction I need a negative indices so I will need -2

    Hence,
    3^2(-2) = 1/81

    There!
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    (Original post by notnek)
    At GCSE, you can't rearrange to solve these.

    Start with 3^x = 81. Can you see what x must be here?

    Then what about 3^x = \frac{1}{81}. Using your answer above, what do you think x must be?

    Finally can you use that to answer your question?

    Please post all your thoughts / working if you get stuck.
    how do u use the different maths symbols?
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    (Original post by Zxrxh)
    how do u use the different maths symbols?
    use "latex" tags to get the different maths symbols...check this website out to see how to use them:
    https://www.thestudentroom.co.uk/wiki/LaTex

    \frac{1}{2}
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    (Original post by NothingButWaleed)
    use "latex" tags to get the different maths symbols...check this website out to see how to use them:
    https://www.thestudentroom.co.uk/wiki/LaTex

    \frac{1}{2}
    To add to this, you can click on the expressions to see what code was used to format them
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    (Original post by NothingButWaleed)
    use "latex" tags to get the different maths symbols...check this website out to see how to use them:
    https://www.thestudentroom.co.uk/wiki/LaTex

    \frac{1}{2}
    thnxx
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    How do you make t the subject of the formula v = u +at.
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    (Original post by Zxrxh)
    How do you make t the subject of the formula v = u +at.
    ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
    \frac{v}{a} = u+t

    then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

    t = \frac{v}{a} - u
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    (Original post by NothingButWaleed)
    ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
    \frac{v}{a} = u+t

    then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

    t = \frac{v}{a} - u
    oh yeah thanks, that was like a grade 3
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    (Original post by Zxrxh)
    oh yeah thanks, that was like a grade 3
    lol yh no worries
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    (Original post by NothingButWaleed)
    ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
    \frac{v}{a} = u+t

    then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

    t = \frac{v}{a} - u
    You didn't divide both sides by a correctly.

    (Original post by Zxrxh)
    oh yeah thanks, that was like a grade 3
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    (Original post by RDKGames)
    You didn't divide both sides by a correctly.
    so what would the answer be ??
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    im failing maths 100%
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    (Original post by Zxrxh)
    so what would the answer be ??
    v=u+at

    Dividing both sides a:

    \frac{v}{a}=\frac{u+at}{a}

    Split the fraction on the RHS into 2 different ones with the same denominator. Simplify where appropriate. Rearrange.

    Have a go.
 
 
 
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