What is the answer to 3^{-x+1}=4^{x+2} ?

The answer to 3^{-x+1}=4^{x+2} is x=-(4ln(2)-ln(3))/(ln(12))

Solve for x: 3^{-x+1}=4^{x+2}

The solution is x=-(4ln(2)-ln(3))/(ln(12))

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3^{- x + 1} = 4^{x + 2}

x = - \frac{4 ln ( 2 ) - ln ( 3 )}{ln ( 12 )}

Decimal Notation

x = - 0.67365 \dots

Solution steps

3^{- x + 1} = 4^{x + 2}

Apply exponent rules

(- x + 1) ln (3) = (x + 2) ln (4)

Solve

(- x + 1) ln (3) = (x + 2) ln (4) : x = - \frac{4 ln ( 2 ) - ln ( 3 )}{ln ( 12 )}

x = - \frac{4 ln ( 2 ) - ln ( 3 )}{ln ( 12 )}

10 e^{8 x + 3} + 2 = 8

3^{x + 4} = 2 7^{2 x - 1}

b^{2^{2 x - 17}} = b^{128}

3^{x - 1} + 3^{x} + 3^{x + 1} = 117

What is the answer to 3^{-x+1}=4^{x+2} ?
The answer to 3^{-x+1}=4^{x+2} is x=-(4ln(2)-ln(3))/(ln(12))
Solve for x: 3^{-x+1}=4^{x+2}
The solution is x=-(4ln(2)-ln(3))/(ln(12))