What is the answer to (12^x)^2+5x-4=12^2x+6 ?

The answer to (12^x)^2+5x-4=12^2x+6 is

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(12^x)^2+5x-4=12^2x+6

(1 2^{x})^{2} + 5 x - 4 = 1 2^{2} x + 6

x = - \frac{139 W _{- 1} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}, x = - \frac{139 W _{0} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}

Decimal Notation

x = 1.00844 \dots, x = - 0.06678 \dots

Solution steps

(1 2^{x})^{2} + 5 x - 4 = 1 2^{2} x + 6

Prepare

(1 2^{x})^{2} + 5 x - 4 = 1 2^{2} x + 6

for Lambert form:

1 = (139 x + 10) e^{- 2 l n (12) x}

Rewrite the equation with

2 ln (12) (- x - \frac{10}{139}) = u

and

x = - \frac{139 u + 20 ln ( 12 )}{278 ln ( 12 )}

1 = (139 (- \frac{139 u + 20 ln ( 12 )}{278 ln ( 12 )}) + 10) e^{- 2 l n (12) (- \frac{139 u + 20 l n ( 12 )}{278 l n ( 12 )})}

Simplify

1 = e^{\frac{139 u + 20 l n ( 12 )}{139}} (- \frac{139 u + 20 ln ( 12 )}{2 ln ( 12 )} + 10)

Rewrite

1 = e^{\frac{139 u + 20 l n ( 12 )}{139}} (- \frac{139 u + 20 ln ( 12 )}{2 ln ( 12 )} + 10)

in Lambert form:

\frac{e ^{\frac{139 u + 20 l n ( 12 )}{139}} u}{e ^{\frac{20 l n ( 12 )}{139}}} = - \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}}

Solve

\frac{e ^{\frac{139 u + 20 l n ( 12 )}{139}} u}{e ^{\frac{20 l n ( 12 )}{139}}} = - \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} : u = W_{- 1} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}}), u = W_{0} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}})

u = W_{- 1} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}}), u = W_{0} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}})

Substitute back

u = 2 ln (12) (- x - \frac{10}{139}),

solve for

x

Solve

2 ln (12) (- x - \frac{10}{139}) = W_{- 1} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}}) : x = - \frac{139 W _{- 1} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}

Solve

2 ln (12) (- x - \frac{10}{139}) = W_{0} (- \frac{2 ln ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}}) : x = - \frac{139 W _{0} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}

x = - \frac{139 W _{- 1} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}, x = - \frac{139 W _{0} ( - \frac{2 l n ( 12 )}{139 e ^{\frac{20 l n ( 12 )}{139}}} ) + 20 ln ( 12 )}{278 ln ( 12 )}

(5^{k})^{2} = \frac{12 5 ^{3}}{( 5 ^{2} )}

2^{2 x + 6 x} - 16 = 0

100 = 5^{b}

0.9711 n - 0.9711 = (0.9614)^{n}

25 6^{2 x - 16} = (\frac{1}{128})^{20 - 2 x}

What is the answer to (12^x)^2+5x-4=12^2x+6 ?
The answer to (12^x)^2+5x-4=12^2x+6 is
Solve for x: (12^x)^2+5x-4=12^2x+6
The solution is