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

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

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

solve for

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

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

Decimal Notation

x = - 0.53286 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

5^{4 x} = 25

1 0^{d} = 10000

e^{x + 2} = 0

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

x, f^{2} = 2^{x - 2}

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