What is the answer to (d^2-9)=(e^3)^x+e^{-3}x ?

The answer to (d^2-9)=(e^3)^x+e^{-3}x is x=-(\W(3/(e^{27e^3-3e^3d^2-3)})-3e^3(d^2-9))/3

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(d^2-9)=(e^3)^x+e^{-3}x

(d^{2} - 9) = (e^{3})^{x} + e^{- 3} x

x = - \frac{W ( \frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}} ) - 3 e ^{3} ( d ^{2} - 9 )}{3}

Solution steps

(d^{2} - 9) = (e^{3})^{x} + e^{- 3} x

Prepare

(d^{2} - 9) = (e^{3})^{x} + e^{- 3} x

for Lambert form:

1 = e^{- 3 x} (d^{2} - 9 - \frac{1}{e ^{3}} x)

Rewrite the equation with

- 3 x + 3 e^{3} (d^{2} - 9) = u

and

x = - \frac{u - 3 e ^{3} ( d ^{2} - 9 )}{3}

1 = e^{- 3 (- \frac{u - 3 e ^{3} ( d ^{2} - 9 )}{3})} (d^{2} - 9 - \frac{1}{e ^{3}} (- \frac{u - 3 e ^{3} ( d ^{2} - 9 )}{3}))

Simplify

1 = e^{u - 3 e^{3} (d^{2} - 9)} (d^{2} - 9 + \frac{u - 3 e ^{3} ( d ^{2} - 9 )}{3 e ^{3}})

Rewrite

1 = e^{u - 3 e^{3} (d^{2} - 9)} (d^{2} - 9 + \frac{u - 3 e ^{3} ( d ^{2} - 9 )}{3 e ^{3}})

in Lambert form:

e^{u} u = \frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}}

Solve

e^{u} u = \frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}} : u = W (\frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}})

u = W (\frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}})

Substitute back

u = - 3 x + 3 e^{3} (d^{2} - 9),

solve for

x

Solve

- 3 x + 3 e^{3} (d^{2} - 9) = W (\frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}}) : x = - \frac{W ( \frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}} ) - 3 e ^{3} ( d ^{2} - 9 )}{3}

x = - \frac{W ( \frac{3}{e ^{27 e^{3} - 3 e^{3} d^{2} - 3}} ) - 3 e ^{3} ( d ^{2} - 9 )}{3}

5^{x} (5^{x} - 4) + 3 = 0

(- 15)^{x} = 1

1 0^{2 x - 3} = (0.1)^{2}

x, 4^{x + y} = 3^{x^{2} + y^{2}}

1 0^{x} = 25

What is the answer to (d^2-9)=(e^3)^x+e^{-3}x ?
The answer to (d^2-9)=(e^3)^x+e^{-3}x is x=-(\W(3/(e^{27e^3-3e^3d^2-3)})-3e^3(d^2-9))/3