Question

log 2 to the base( 3x^2 + 1) < (1/2)
Find out the solution of this inequality.

Answer 1

$log_{3 {x}^{2} + 1}(2) < \frac{1}{2}$

$2 < {(3x {}^{2} + 1) }^{ \frac{1}{2} }$

$4 < 3x {}^{2} + 1$

$3 < 3 {x}^{2}$

${x}^{2} > 1$

$x \: is \: in \: ( - \infty \: - 1) \: u \: (1 \: \infty )$

Answer 2

Step-by-step explanation:

log

3x

2

+1

(2)<

2

1

2 < {(3x {}^{2} + 1) }^{ \frac{1}{2} }2<(3x

2

+1)

2

1

4 < 3x {}^{2} + 14<3x

2

+1

3 < 3 {x}^{2}3<3x

2

{x}^{2} > 1x

2

>1

x \: is \: in \: ( - \infty \: - 1) \: u \: (1 \: \infty )xisin(−∞−1)u(1∞)