Question

solve the inequality log (2+x) to the base x square less than 1​

Answer 1

it is given that $log_{x^2}(2+x)<1$

for log to be defined,

(2 + x) > 0 ⇒x > -2 .......(1)

x² > 0 for all x belongs to R.

and x² ≠ 1 ⇒ x ≠ 1

case 1 : if we take x² > 1 ⇒x > 1 or, x < -1

now, $log_{x^2}(2+x)<1$

⇒ (2 + x) < x²

⇒ x² - x - 2 < 0

⇒ x² - 2x + x - 2 < 0

⇒ x(x - 2) + 1(x - 2) < 0

⇒ (x + 1)(x - 2) < 0

⇒x > 2 or, x < -1

taking common inequality from x >1 or, x < -1 and x > 2 or, x < -1

we get, x > 2 or x < -1

so, x $\in$ (2, ∞) U (-∞, -1) .....(2)

from equations (1) and (2),

$x\in$ (2, ∞) U (-2, -1) .......(3)

case 2 : if we take 0 < x² < 1 ⇒ -1 < x < 1, x ≠ 0

$log_{x^2}(2+x)<1$

or, (2 + x) > x²

or, x² - x - 2 < 0

or, (x - 2)(x + 1) < 0

or, -1 < x < 2

taking common inequality from -1 < x < 1 ,x ≠ 0 and -1 < x < 2

or, x $\in$(-1, 0) U(0, 1) ......(4)

taking common inequality of equations (3) and (4),

we get, x$\in$(-2, -1) U (-1, 0) U (0, 1) U (2, ∞)

Answer 2

Step-by-step explanation:

it is given that log_{x^2}(2+x) < 1logx2(2+x)<1

for log to be defined,

(2 + x) > 0 ⇒x > -2 .......(1)

x² > 0 for all x belongs to R.

and x² ≠ 1 ⇒ x ≠ 1

case 1 : if we take x² > 1 ⇒x > 1 or, x < -1

now, log_{x^2}(2+x) < 1logx2(2+x)<1

⇒ (2 + x) < x²

⇒ x² - x - 2 < 0

⇒ x² - 2x + x - 2 < 0

⇒ x(x - 2) + 1(x - 2) < 0

⇒ (x + 1)(x - 2) < 0

⇒x > 2 or, x < -1

taking common inequality from x >1 or, x < -1 and x > 2 or, x < -1

we get, x > 2 or x < -1

so, x \in∈ (2, ∞) U (-∞, -1) .....(2)

from equations (1) and (2),

x\inx∈ (2, ∞) U (-2, -1) .......(3)

case 2 : if we take 0 < x² < 1 ⇒ -1 < x < 1, x ≠ 0

log_{x^2}(2+x) < 1logx2(2+x)<1

or, (2 + x) > x²

or, x² - x - 2 < 0

or, (x - 2)(x + 1) < 0

or, -1 < x < 2

taking common inequality from -1 < x < 1 ,x ≠ 0 and -1 < x < 2

or, x \in∈ (-1, 0) U(0, 1) ......(4)

taking common inequality of equations (3) and (4),

we get, x\in∈ (-2, -1) U (-1, 0) U (0, 1) U (2, ∞)