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Thread: Simultaneous equations (tough one): 8^x 64^y = 128, sqrt[12y - 7] = x

  1. #11
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Denis View Post
    8^x * 64^y = 128 [1]

    SQRT(12y - 7) = x [2]

    By inspection of [1], easily seen that x=1/3 and y=1.
    Other solution not easily seen?

    Find it, by first solving for y: 18y^2 - 96y + 56 = 0
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  2. #12
    Quote Originally Posted by Denis View Post
    8^x * 64^y = 128 [1]
    SQRT(12y - 7) = x [2]

    By inspection of [1], easily seen that x=1/3 and y=1.

    Seems that [2] contains a typo(s)....
    No ,mister, no typos

  3. #13
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    Rewrite Equation 1

    2^(3x+6y) = 2^7

    Then equation 1 becomes simply,
    3x+6y = 7
    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

  4. #14
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    Quote Originally Posted by mmm4444bot View Post
    Other solution not easily seen?
    According to 1st post, the OP apparently was learning
    how to solve "2 unknowns, 2 equations" problems.

    I showed one possible solution to his 1st equation.
    This was mainly to see if the OP understood.

    I made up a similar case:
    x=3, y=2
    2^x * 5^y = 200 [1]
    sqrt(12y - 15) = x [2]

    Then used Wolfram (to "see"):
    http://www.wolframalpha.com/input/?i...t(12*y-15)%3Dx

    So no problems with a "proper" set of 2 equations.

    I then tried the OP's 2 equations:
    http://www.wolframalpha.com/input/?i...rt(12*y-7)%3Dx

    No results...Wolfram gave up!

    So I assumed there was something wrong
    with the equations...

    My apologies...
    I'm just an imagination of your figment !

  5. #15
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Denis View Post
    I showed one possible solution to his 1st equation.
    This was mainly to see if the OP understood.
    Did you see their work?


    Then used Wolfram ... No results ...Wolfram gave up!
    When I tried it, Wolfram showed the solution graphically. If you want to enlarge the graph or see further computations, ya gotta pay.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  6. #16
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    Quote Originally Posted by Bob Brown MSEE View Post
    2^(3x+6y) = 2^7

    Then equation 1 becomes simply,
    3x+6y = 7
    6x+12y = 14 [1]
    12y - 7 = xx [2]

    so
    6x+ (xx+7) = 14

    x=1 or x=-7

    trying in [1], x=1
    Answ: x=1, y=2/3

    Wolfram Solution
    Last edited by Bob Brown MSEE; 06-20-2017 at 05:03 AM.
    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

  7. #17
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    I'm just an imagination of your figment !

  8. #18
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Bob Brown MSEE View Post
    6x+12y = 14 [1]
    12y - 7 = xx [2]

    so
    6x+ (xx+7) = 14
    I like it.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  9. #19
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Tuharramah View Post
    Sqrt (12y-7)=x
    8^x * 64^y=128

    So 2^3x + 2^6y = 2^7

    3*sqrt (12y-7) +6y =7
    I wouldn't go this way. Instead:

    . . . . .3x + 6y = 7

    . . . . .12y - 7 = x^2

    Multiply the first equation by -2:

    . . . . .-6x - 12y = -14

    . . . . .12y - 7 = x^2

    Add down to get:

    . . . . .-6x - 7 = x^2 - 14

    . . . . .0 = x^2 + 6x - 7

    . . . . .0 = (x + 7)(x - 1)

    ...and so forth.

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