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Math Help - INDICES PROBLEM SOLVING

  1. #1
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    INDICES PROBLEM SOLVING

    X = 4 10m Y = 2 10n
    Where m and n are integers.
    XY= 8 102 and x/y = 2 108
    Find the value of m and the value of n
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  2. #2
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    Quote Originally Posted by abey_27 View Post
    X = 4 10m Y = 2 10n
    Where m and n are integers.

    XY= 8 102 and x/y = 2 108
    Find the value of m and the value of n
    XY = 8 \times 102

    (4 \times 10m)(2 \times 10n) = 8 \times 102

    8 \times 100 \times mn = 8 \times 102

    mn = \frac{102}{100} .........eqn(1)

    \frac{X}{Y} = 2 \times 108

    \frac{4 \times 10m}{2 \times 10n} = 2 \times 108

    \frac{2m}{n} = 2 \times 108

    \frac{m}{n} =108

    m = 108 n ..............eqn(2)

    Put this value of m in eqn (1),

    (108n)n = \frac{102}{100}

    108n^2 = \frac{102}{100}

    n^2 = \frac{102}{100 \times 108}

    n= \sqrt{\frac{102}{100 \times 108}} , now, simplify to get integer value

    put this value of n in eqn(1),

    m = 108 \times \sqrt{\frac{102}{100 \times 108}}

    m = \sqrt{\frac{108 \times 102}{100}}, now, simplify to get integer value
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  3. #3
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    sorry shyam, i forgot to put the power indices symbol on, could you possibly do the Q again, much appreciated thanks

    X = 4 10^m Y = 2 10^n
    Where m and n are integers.
    XY= 8 10^2 and x/y = 2 10^8
    Find the value of m and the value of n
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  4. #4
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    Solution involves simultaneous equations.

    x=4X10^m and y=2X10^n

    xy=8X10^2

    hence (4X10^m)(2X10^n) = 8X10^2
    so 8X10^(m+n) = 8X10^2
    simplified: 10^(m+n) = 10^2 (divide by 8)
    hence m+n = 2

    x/y=2X10^8

    hence (4X10^m)/(2X10^n) = 2X10^8
    so 2X10^(m-n) = 2X10^8
    simplified: 10^(m-n) = 10^8 (divide by 2)
    hence m-n = 8


    now m+n = 2
    + m-n = 8
    _____________

    2m = 10 (since n-n = 0)

    hence m = 5 (since 2X5 = 10)

    substitute into m+n = 2
    5+n = 2
    n = 2 - 5
    n = -3

    so: m=5 and n=-3

    Lovely question. Hope this helps. Cheers.
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  5. #5
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    thank you so much, could you possibly try the other indice question as well, highly appreciated
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  6. #6
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    Quote Originally Posted by abey_27 View Post
    X = 4 10^m Y = 2 10^n
    Where m and n are integers.
    XY= 8 10^2 and x/y = 2 10^8
    Find the value of m and the value of n
    XY = 8 \times 10^{n+m} and X/Y = 2 \times 10^{n-m}.

    This means n+m = 2 and n-m=8.
    Can you solve these equations?
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  7. #7
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    yea, thanks you for your help. I have just posted another algebra Q, if you have time, could you possibly help me on that one. thank you again
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  8. #8
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    Quote Originally Posted by abey_27 View Post
    sorry shyam, i forgot to put the power indices symbol on, could you possibly do the Q again, much appreciated thanks

    X = 4 10^m Y = 2 10^n
    Where m and n are integers.
    XY= 8 10^2 and x/y = 2 10^8
    Find the value of m and the value of n

    XY = 8 \times 10^2

    (4 \times 10^m)(2 \times 10^n) = 8 \times 10^2

    8 \times 10^{m+n} = 8 \times 10^2

    10^{m+n} = 10^2

    m+n = 2 .........eqn(1)

    Now,

    \frac{X}{Y} = 2 \times 10^8

    \frac{4 \times 10^m}{2 \times 10^n} = 2 \times 10^8

    2\times 10^{m-n}= 2 \times 10^8

    10^{m-n}=10^8

    m-n=8 ................eqn(2)

    Adding equations (1) and (2)

    m+n = 2
    m-n=8
    -----------------------------------
    2m = 10
    -----------------------------------

    m = 5

    Put this value of m in eqn(1),

    5 + n = 2

    n = - 3

    so, m = 5 and n = -3

    Did you get it now???
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