Question

the values of x(where x>1)satisfying logx(x^3-x^2-2x)<3 is

Answer 1

We have to find the value of x satisfying $log_x^{(x^3-x^2-2x)}$ < 3

solution : it has given that, $log_x^{(x^3-x^2-2x)}$ < 3

⇒log(x³ - x² - 2x)/logx < 3

⇒{log(x³ - x² - 2x) - 3logx }/logx < 0

⇒{log(x³ - x² - 2x) - logx³}/log x < 0

⇒log [(x³ - x² - 2x)/x³ ]/logx < 0

⇒log[x(x² - x - 2)/x³]/logx < 0

⇒log[(x - 2)(x + 1)/x² ]/logx < 0

⇒$log_x^{(x-2)(x+1)/x^2}$ < 0

x > 0 but x ≠ 1 .....(1)

and 0 < (x - 2)(x + 1)/x² < 1

case 1 : ⇒(x - 2)(x + 1)/x² > 0

⇒x > 2, x < -1 ......(2)

case 2 : (x - 2)(x + 1)/x² < 1

⇒(x² - x - 2 - x²)/x² < 0

⇒-(x + 2)/x² < 0

⇒(x + 2)/x² > 0

⇒x > -2 ....(3)

from equations (1), (2) and (3) we get,

x > 2

Therefore the value of x for which expression is satisfying, is x > 2 i.e., (2, ∞)

Answer 2

x∈ (2,∞) Answer.....................