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

The answer to 4^{1-x}=3^{2x+4} is x=-(4ln(3)-2ln(2))/(2ln(6))

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

The solution is x=-(4ln(3)-2ln(2))/(2ln(6))

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

x = - \frac{4 ln ( 3 ) - 2 ln ( 2 )}{2 ln ( 6 )}

Decimal Notation

x = - 0.83944 \dots

Solution steps

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

Apply exponent rules

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

Solve

(1 - x) ln (4) = (2 x + 4) ln (3) : x = - \frac{4 ln ( 3 ) - 2 ln ( 2 )}{2 ln ( 6 )}

x = - \frac{4 ln ( 3 ) - 2 ln ( 2 )}{2 ln ( 6 )}

2 7^{3 x + 5} = 8 1^{4 x + 1}

4 = 102 4^{x}

0.2 5^{2 x - 1} = 8^{x + 3}

86 \cdot 1 0^{\frac{x}{4}} = 89

3^{4 x - 7} = \frac{3 ^{2 x}}{3}

What is the answer to 4^{1-x}=3^{2x+4} ?
The answer to 4^{1-x}=3^{2x+4} is x=-(4ln(3)-2ln(2))/(2ln(6))
Solve for x: 4^{1-x}=3^{2x+4}
The solution is x=-(4ln(3)-2ln(2))/(2ln(6))