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

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?