Question

Solve the inequation

${3}^{x + 2} > {( \frac{1}{9}) }^{ \frac{1}{x} }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequation is

$\rm \: {3}^{x + 2} > {\bigg(\dfrac{1}{9} \bigg) }^{\dfrac{1}{x} }$

can be rewritten as

$\rm \: {3}^{x + 2} > {\bigg(\dfrac{1}{ {3}^{2} } \bigg) }^{\dfrac{1}{x} }$

$\rm \: {3}^{x + 2} > {\bigg( {3}^{ - 2} \bigg) }^{\dfrac{1}{x} }$

$\rm \: {3}^{x + 2} > {\bigg(3\bigg) }^{\dfrac{ - 2}{x} }$

$\rm\implies \:x + 2 > - \dfrac{2}{x}$

$\rm \: x + 2 + \dfrac{2}{x} > 0$

$\rm \: \dfrac{ {x}^{2} + 2x + 2}{x} > 0$

Now, x² + 2x + 2 is a quadratic polynomial such that coefficient of x² = 1 > 0 and Discriminant, D = 4 - 8 = - 4 < 0

So, we know, in a quadratic expression ax² + bx + c, if a > 0 and D < 0, then ax² + bx + c > 0

$\rm\implies \: {x}^{2} + 2x + 2 > 0$

Thus,

$\rm\implies \:\dfrac{1}{x} > 0$

$\rm\implies \:x > 0$

$\rm\implies \:x \in \: (0, \: \infty )$

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ADDITIONAL INFORMATION

$\boxed{\tt{ |x| < y \: \: \rm\implies \: - y < x < y \: }}$

$\boxed{\tt{ |x| \leqslant y \: \: \rm\implies \: - y \leqslant x \leqslant y \: }}$

$\boxed{\tt{ |x - z| \leqslant y \: \: \rm\implies \:z - y \leqslant x \leqslant z + y \: }}$

$\boxed{\tt{ |x - z| < y \: \: \rm\implies \:z - y < x < z + y \: }}$

$\boxed{\tt{ |x| > y \: \: \rm\implies \:x < - y \: \: or \: \: x > y \: }}$

$\boxed{\tt{ |x| \geqslant y \: \: \rm\implies \:x \leqslant - y \: \: or \: \: x \geqslant y \: }}$

Answer 2

Step-by-step explanation:

Solution−

Given inequation is

\begin{gathered}\rm \: {3}^{x + 2} > {\bigg(\dfrac{1}{9} \bigg) }^{\dfrac{1}{x} } \\ \end{gathered}

3

x+2

>(

9

1

)

x

1

can be rewritten as

\begin{gathered}\rm \: {3}^{x + 2} > {\bigg(\dfrac{1}{ {3}^{2} } \bigg) }^{\dfrac{1}{x} } \\ \end{gathered}

3

x+2

>(

3

2

1

)

x

1

\begin{gathered}\rm \: {3}^{x + 2} > {\bigg( {3}^{ - 2} \bigg) }^{\dfrac{1}{x} } \\ \end{gathered}

3

x+2

>(3

−2

)

x

1

\begin{gathered}\rm \: {3}^{x + 2} > {\bigg(3\bigg) }^{\dfrac{ - 2}{x} } \\ \end{gathered}

3

x+2

>(3)

x

−2

\rm\implies \:x + 2 > - \dfrac{2}{x}⟹x+2>−

x

2

\rm \: x + 2 + \dfrac{2}{x} > 0x+2+

x

2

>0

\rm \: \dfrac{ {x}^{2} + 2x + 2}{x} > 0

x

x

2

+2x+2

>0

Now, x² + 2x + 2 is a quadratic polynomial such that coefficient of x² = 1 > 0 and Discriminant, D = 4 - 8 = - 4 < 0

So, we know, in a quadratic expression ax² + bx + c, if a > 0 and D < 0, then ax² + bx + c > 0

\rm\implies \: {x}^{2} + 2x + 2 > 0⟹x

2

+2x+2>0

Thus,

\rm\implies \:\dfrac{1}{x} > 0⟹

x

1

>0

\rm\implies \:x > 0⟹x>0

\rm\implies \:x \in \: (0, \: \infty )⟹x∈(0,∞)