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

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

Solve for x: 3*2^{-x}=2*3^x

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

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3 \cdot 2^{- x} = 2 \cdot 3^{x}

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

Decimal Notation

x = 0.22629 \dots

Solution steps

3 \cdot 2^{- x} = 2 \cdot 3^{x}

Apply exponent rules

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

Solve

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

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

9^{x} = 3^{x + 3} - 5

85 = 1 1^{x}

5^{x + 1} = 7^{5 x - 1}

- 4 = (\frac{1}{2})^{x}

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