9 to the power x = 27 to the power y, 64 to the powerxy = 152 to the power x+1?

solve the simutaneous equation

6 Answers

  • 9^x = 27^y

    becomes

    (3^2)^x = (3^3)^y

    3^(2x) = 3^(3y)

    so…

    2x = 3y

    y = 2x/3

    64^(xy) = 152^(x+1)

    (2^6)^(xy) = (152^x)(152)

    2^(6xy) = 152(152^x)

    ln 2^(6xy) = ln 152(152^x)

    6xy (ln 2) = ln 152 + ln 152^x

    6xy (ln 2) = ln 152 + x ln 152

    Plug in y = 2x/3

    6x(2x/3) (ln 2) = ln 152 + x ln 152

    4x^2 ln 2 – x ln 152 – ln 152 = 0

    using quadratic formula…

    a = 4 ln 2

    b = -ln 152

    c = -ln 152

    This becomes quite a mess. Do you have all the information typed correctly?

    ============================

    Did you mean 512, instead of 152?

    64^xy = 512^(x+1)

    (2^6)^xy = (2^9)^(x+1)

    2^(6xy) = 2^(9(x+1))

    so

    6xy = 9(x+1)

    Now you have two simpler equations…

    y = 2x/3

    and

    6xy = 9(x + 1)

    solve by substitution…

    6x(2x/3) = 9x + 9

    4x^2 = 9x + 9

    4x^2 – 9x – 9 = 0

    (4x + 3)(x – 3) = 0

    x = -3/4 and 3

    Plug those into y = 2x/3 to find y…

    so…

    (3, 2) and (-3/4 , -1/2)

    are the two answers.

  • If you put 9 and 27 as powers of 3 you will have:

    (3^2)^x=(3^3)^y Which you can then turn into:

    3^2x=3^3y And then using like bases, you have:

    2x=3y Where you can solve for x (because solving for y would give you a yucky fraction) and you have:

    x=1.5y

    If you put 64 and 152 as powers of…hmmm, they don’t have a common base. Okay, replace the x’s with the formula we got above. This would give you:

    64^(1.5y^2) = 152^(1.5y+1) Then take the logs of both sides, this would give you:

    (1.5y^2)(log64)=(1.5y+1)(log 152) Subtract to get 0 on right side of eqn, and you would get:

    (1.5y^2)(log64)-(1.5y+1)(log 152)=0

    Then solve with graphing calculator or even quadratic formula.

    With a graphing calc, I got the answers:

    y = -.4777334 & y = 1.6857213

    And then find x by multiplying y by 1.5 and you have the corresponding x vaules of:

    x=-.7166001 & x=2.52858195

  • if 9^x = 27^y… You can do a logarithm of both sides:

    xln9 = yln27

    ln9 = ln3^2=2ln3, ln27 = ln3^3=3ln3

    so y =2/3x

    64^xy = 152^(x+1) then

    xyln64 = (x+1)ln152. since y =2/3x

    2/3x^2ln64 = (x+1)ln152. 64=2^6 so ln64 = 6ln2. Your equation becomes:

    4ln2*x^2 -ln152*x -ln152 = 0

    Now I am assuming you meant 512 and not 152.

    ln512 = ln2^9 = 9ln2.

    Your equation becomes :

    4ln2*x^2 -9ln2*x -9ln2 = 0 , divide by ln2:

    4x^2 -9x -9 = 0 which is solved as :

    (x-3)(4x+3)=0 :x=3 or x =-3/4.

    since a^x is only possible for positive values of x, then x=3

    y=2/3x, so y=2

  • 9 X 27

  • 9^x=27^y => 3^2x=3^3y => 2x=3y =>6y=4x.

    64^xy=152^(x+1) (!?) =>2^(6yx)=152^(x+1) =>

    2^(4x^2)=152^(x+1) and … solution must be determine roughly!

  • Is the question correct?

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