What is the answer to (5^x)^2+8=125^2x ?

The answer to (5^x)^2+8=125^2x is

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(5^x)^2+8=125^2x

(5^{x})^{2} + 8 = 12 5^{2} x

x = - \frac{15625 W _{- 1} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}, x = - \frac{15625 W _{0} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}

Decimal Notation

x = 3.37813 \dots, x = 0.00057 \dots

Solution steps

(5^{x})^{2} + 8 = 12 5^{2} x

Prepare

(5^{x})^{2} + 8 = 12 5^{2} x

for Lambert form:

1 = (15625 x - 8) e^{- 2 l n (5) x}

Rewrite the equation with

2 ln (5) (- x + \frac{8}{15625}) = u

and

x = - \frac{15625 u - 16 ln ( 5 )}{31250 ln ( 5 )}

1 = (15625 (- \frac{15625 u - 16 ln ( 5 )}{31250 ln ( 5 )}) - 8) e^{- 2 l n (5) (- \frac{15625 u - 16 l n ( 5 )}{31250 l n ( 5 )})}

Simplify

1 = e^{\frac{15625 u - 16 l n ( 5 )}{15625}} (- \frac{15625 u - 16 ln ( 5 )}{2 ln ( 5 )} - 8)

Rewrite

1 = e^{\frac{15625 u - 16 l n ( 5 )}{15625}} (- \frac{15625 u - 16 ln ( 5 )}{2 ln ( 5 )} - 8)

in Lambert form:

e^{u} u = - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625}

Solve

e^{u} u = - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625} : u = W_{- 1} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625}), u = W_{0} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625})

u = W_{- 1} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625}), u = W_{0} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625})

Substitute back

u = 2 ln (5) (- x + \frac{8}{15625}),

solve for

x

Solve

2 ln (5) (- x + \frac{8}{15625}) = W_{- 1} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625}) : x = - \frac{15625 W _{- 1} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}

Solve

2 ln (5) (- x + \frac{8}{15625}) = W_{0} (- \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} ln ( 5 )}{15625}) : x = - \frac{15625 W _{0} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}

x = - \frac{15625 W _{- 1} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}, x = - \frac{15625 W _{0} ( - \frac{2 e ^{\frac{16 l n ( 5 )}{15625}} l n ( 5 )}{15625} ) - 16 ln ( 5 )}{31250 ln ( 5 )}

15 - 0. 5^{x} = - 6

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64 = 2^{x}

2 = (1.1)^{x}

x, y^{12} = y^{2 x}