Question 37379: 9^x+4=27^x^2
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! From 9^(x+4) = 27^(x^2), we turn both the 9 and the 27 into powers of three:
(3^2)^(x + 4) = (3^3)^(x^2)
Now use the laws of exponents...
3^(2x + 8) = 3^(3x^2)
Now equate the exponents...
2x + 8 = 3x^2
and solve this...
3x^2 - 2x - 8 = 0
(3x + 4)(x - 2) = 0
x = -4/3 or x = 2
|
|
|
