What is ((12)^x)^2+4x-3>= 9 ?

The solution to ((12)^x)^2+4x-3>= 9 is x>=-(\W_{0}(1492992ln(12)))/(2ln(12))+3

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((12)^x)^2+4x-3>= 9

((12)^{x})^{2} + 4 x - 3 \geq 9

x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

Interval Notation

[- \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3, \infty)

Decimal Notation

x \geq 0.46602 \dots

Solution steps

((12)^{x})^{2} + 4 x - 3 \geq 9

Apply exponent rule:

(a^{b})^{c} = a^{b c}

1 2^{2 x} + 4 x - 3

Move

4 x

to the right side

1 2^{2 x} - 3 \geq 9 - 4 x

Move

3

to the right side

1 2^{2 x} \geq - 4 x + 12

Prepare

1 2^{2 x} \geq - 4 x + 12

for Lambert form:

1 \geq e^{- 2 l n (12) x} (- 4 x + 12)

Rewrite the equation with

2 ln (12) (- x + 3) = u

and

x = - \frac{u - 6 ln ( 12 )}{2 ln ( 12 )}

1 \geq e^{- 2 l n (12) (- \frac{u - 6 l n ( 12 )}{2 l n ( 12 )})} (- 4 (- \frac{u - 6 ln ( 12 )}{2 ln ( 12 )}) + 12)

Simplify

1 \geq e^{u - 6 l n (12)} (\frac{2 ( u - 6 ln ( 12 ) )}{ln ( 12 )} + 12)

Rewrite

1 \geq e^{u - 6 l n (12)} (\frac{2 ( u - 6 ln ( 12 ) )}{ln ( 12 )} + 12)

in Lambert form:

e^{u} u \leq 1492992 ln (12)

Solve

e^{u} u \leq 1492992 ln (12) : u \leq W_{0} (1492992 ln (12))

u \leq W_{0} (1492992 ln (12))

Substitute back

2 ln (12) (- x + 3) = u,

solve for

x

Solve

2 ln (12) (- x + 3) \leq W_{0} (1492992 ln (12)) : x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

x \geq - \frac{W _{0} ( 1492992 ln ( 12 ) )}{2 ln ( 12 )} + 3

1 + e^{x} > 0

(0.3)^{2 x^{2} - 3 x - 6} < 0.00243

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

5^{1 - x} < 5^{4 - 5 x}

x, (\frac{3}{2})^{1 + r \cdot l^{2}} > (\frac{2}{3})^{3 x - 1}

What is ((12)^x)^2+4x-3>= 9 ?
The solution to ((12)^x)^2+4x-3>= 9 is x>=-(\W_{0}(1492992ln(12)))/(2ln(12))+3