What is the answer to 9-3x=(1/27)^{x+3} ?

The answer to 9-3x=(1/27)^{x+3} is x=-(-\W_{0}(-\frac{ln(3))/(387420489))-9ln(3)}{3ln(3)}

Solve for x: 9-3x=(1/27)^{x+3}

The solution is x=-(-\W_{0}(-\frac{ln(3))/(387420489))-9ln(3)}{3ln(3)}

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9 - 3 x = (\frac{1}{27})^{x + 3}

x = - \frac{- W _{0} ( - \frac{l n ( 3 )}{387420489} ) - 9 ln ( 3 )}{3 ln ( 3 )}

Decimal Notation

x = 2.99999 \dots

Solution steps

9 - 3 x = (\frac{1}{27})^{x + 3}

Prepare

9 - 3 x = (\frac{1}{27})^{x + 3}

for Lambert form:

(9 - 3 x) e^{- 3 l n (3) (- x - 3)} = 1

Rewrite the equation with

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

and

x = - \frac{- u - 9 ln ( 3 )}{3 ln ( 3 )}

(9 - 3 (- \frac{- u - 9 ln ( 3 )}{3 ln ( 3 )})) e^{- 3 l n (3) (- (- \frac{- u - 9 l n ( 3 )}{3 l n ( 3 )}) - 3)} = 1

Simplify

e^{u + 18 l n (3)} (9 + \frac{- u - 9 ln ( 3 )}{ln ( 3 )}) = 1

Rewrite

e^{u + 18 l n (3)} (9 + \frac{- u - 9 ln ( 3 )}{ln ( 3 )}) = 1

in Lambert form:

e^{u} u = - \frac{ln ( 3 )}{387420489}

Solve

e^{u} u = - \frac{ln ( 3 )}{387420489} : u = W_{0} (- \frac{ln ( 3 )}{387420489})

u = W_{0} (- \frac{ln ( 3 )}{387420489})

Substitute back

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

solve for

x

Solve

- 3 ln (3) (- x + 3) = W_{0} (- \frac{ln ( 3 )}{387420489}) : x = - \frac{- W _{0} ( - \frac{l n ( 3 )}{387420489} ) - 9 ln ( 3 )}{3 ln ( 3 )}

x = - \frac{- W _{0} ( - \frac{l n ( 3 )}{387420489} ) - 9 ln ( 3 )}{3 ln ( 3 )}

9^{x} - 43 \cdot 3^{x - 1} = 0

x, y = e^{\frac{x}{b}}

9^{x} = 81 \cdot 3^{x + 2}

e^{9 x - 1} = 6

8^{2 x + 1} = 128

What is the answer to 9-3x=(1/27)^{x+3} ?
The answer to 9-3x=(1/27)^{x+3} is x=-(-\W_{0}(-\frac{ln(3))/(387420489))-9ln(3)}{3ln(3)}
Solve for x: 9-3x=(1/27)^{x+3}
The solution is x=-(-\W_{0}(-\frac{ln(3))/(387420489))-9ln(3)}{3ln(3)}