SOLUTION: (1/3)^2-x=27^3+x
Algebra.Com
Question 181203: (1/3)^2-x=27^3+x
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(1/3)^2-x=27^3+x
-------------------
I'll assume the exponents are (2-x) and (3+x).
(1/3)^(2-x) = 3^(x-2)
And, 27^(3+x) = 3^(9+3x) (27 = 3^3)
So,
3^(x-2) = 3^(3x+9)
x-2 = 3x+9
2x = -11
x = -11/2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(1/3)^(2-x)=27^(3+x)
-----------------------
3^(-(2-x)) = 3^(3(3+x))
3^(-2+x) = 3^(9+3x)
-2+x = 9+3x
2x = -11
x = -11/2
=====================
Cheers,
Stan H.
RELATED QUESTIONS
Solve: 3^(x^2+4x) =... (answered by stanbon)
3^(x^2+20)=(1/27)^(3x) (answered by nycsub_teacher)
... (answered by Fombitz)
x^(3/2)... (answered by mananth)
(x^2+27)+(x+3) (answered by Alan3354)
81^(X-2)(1/27^(-X-1))=9^(2X-3) (answered by Tatiana_Stebko)
27/(x+1)^2-3/(x+1)=4 (answered by math33)
log ^27=3... (answered by Alan3354)
3^x+1=27/x^3 (answered by Alan3354)