solve the simutaneous equation
9^x = 27^y
(3^2)^x = (3^3)^y
3^(2x) = 3^(3y)
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))
6xy = 9(x+1)
Now you have two simpler equations…
y = 2x/3
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…
(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:
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:
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?