What is the answer to 3^{x+2}=2^{1-x} ?

The answer to 3^{x+2}=2^{1-x} is x=(ln(2)-2ln(3))/(ln(6))

Solve for x: 3^{x+2}=2^{1-x}

The solution is x=(ln(2)-2ln(3))/(ln(6))

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

x = \frac{ln ( 2 ) - 2 ln ( 3 )}{ln ( 6 )}

Decimal Notation

x = - 0.83944 \dots

Solution steps

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

Apply exponent rules

(x + 2) ln (3) = (1 - x) ln (2)

Solve

(x + 2) ln (3) = (1 - x) ln (2) : x = \frac{ln ( 2 ) - 2 ln ( 3 )}{ln ( 6 )}

x = \frac{ln ( 2 ) - 2 ln ( 3 )}{ln ( 6 )}

e^{x} = x^{10}

e^{2 x} = 18

(e^{4})^{x} (e^{x})^{2} = e^{12}

65 = 1 9^{x}

e^{1 - 4 x} = 2

What is the answer to 3^{x+2}=2^{1-x} ?
The answer to 3^{x+2}=2^{1-x} is x=(ln(2)-2ln(3))/(ln(6))
Solve for x: 3^{x+2}=2^{1-x}
The solution is x=(ln(2)-2ln(3))/(ln(6))