What is the answer to 3^{1-2x}=4^{2x+5} ?

The answer to 3^{1-2x}=4^{2x+5} is x=-(10ln(2)-ln(3))/(2ln(12))

Solve for x: 3^{1-2x}=4^{2x+5}

The solution is x=-(10ln(2)-ln(3))/(2ln(12))

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3^{1 - 2 x} = 4^{2 x + 5}

x = - \frac{10 ln ( 2 ) - ln ( 3 )}{2 ln ( 12 )}

Decimal

x = - 1.17365 \dots

Solution steps

3^{1 - 2 x} = 4^{2 x + 5}

Apply exponent rules

(1 - 2 x) ln (3) = (2 x + 5) ln (4)

Solve

(1 - 2 x) ln (3) = (2 x + 5) ln (4) : x = - \frac{10 ln ( 2 ) - ln ( 3 )}{2 ln ( 12 )}

x = - \frac{10 ln ( 2 ) - ln ( 3 )}{2 ln ( 12 )}

3^{4 x} - 3^{2 x} - 6 = 0

(- 3)^{x} = (- 3)^{13}

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

3^{x} = 7^{x}

y, \frac{1}{3 ^{- 4}} = x^{y}

What is the answer to 3^{1-2x}=4^{2x+5} ?
The answer to 3^{1-2x}=4^{2x+5} is x=-(10ln(2)-ln(3))/(2ln(12))
Solve for x: 3^{1-2x}=4^{2x+5}
The solution is x=-(10ln(2)-ln(3))/(2ln(12))