Question

Solve the inequality for x. Please post whole solution.​

Answer 1

Given:-

$log_{x} ( {a}^{2} + 1) < 0$

To Find:-

• x.

Solution:-

We Know That Result Of Even Power will always Be Positive

So,

a² is always be Positive and a² will never be equal to 0.

$log_{x} ( {a}^{2} + 1) < 0$

OR

we can write it as

$\frac{ log ( {a}^{2} + 1) }{ log \: x} < 0$

Now,

$log \: x < 0$

As

$log( {a}^{2} + 1) < 0$will always be positive

$\therefore \: 0 < x < 1$

Answer 2

Answer:

Given:-

log_{x} ( {a}^{2} + 1) < 0log

x

(a

2

+1)<0

To Find:-

x.

Solution:-

We Know That Result Of Even Power will always Be Positive

So,

a² is always be Positive and a² will never be equal to 0.

log_{x} ( {a}^{2} + 1) < 0log

x

(a

2

+1)<0

OR

we can write it as

\frac{ log ( {a}^{2} + 1) }{ log \: x} < 0

logx

log(a

2

+1)

<0

Now,

log \: x < 0logx<0

As

log( {a}^{2} + 1) < 0log(a

2

+1)<0 will always be positive

\therefore \: 0 < x < 1∴0<x<1

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