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

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?

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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?