Question

What is value of x in the equation(x+ 4)^5 =3125

Answer 1

(x + 4)^5 = 3125
(x + 4)(x + 4)(x + 4)(x + 4)(x + 4) = 3125
(x² + 4x + 4x + 16)(x² + 4x + 4x + 16)(x + 4) = 3125
(x² + 8x + 16)(x² + 8x + 16)(x + 4) = 3125
(x^4 + 8x³ + 16x² + 8x³ + 64x + 128x + 16x² + 128x + 256)(x + 4) = 3125
(x^4 + 8x³ + 8x³ + 16x² + 16x² + 64x + 128x + 128x + 256)(x + 4) = 3125
(x^4 + 16x³ + 32x² + 320x + 256)(x + 4) = 3125
x^5 + 4x^4 + 16x^4 + 64x³ + 32x³ + 128x² + 320x² + 1280x + 256x + 1024 = 3125
x^5 + 20x^4 + 96x³ + 448x² + 1536x + 1024 = 3125
                                                             -1024   -1024
            x^5 + 20x^4 + 96x³ + 448x² + 1536x = 2101
                                                                     x = 1

Answer 2

(X+4)^5=3125
Use distributive property for x and 5 and 4 and 5 to get
5x+20=3125
Then subtract 20 from both sides to get
5x=3105
Then divide both sides by 5 to get
X= 621