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

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

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

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

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

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

Solution steps

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

Apply exponent rules

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

Solve

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

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

9^{x} - 3^{x + 2} + 18 = 0

1.4 \cdot 0. 9^{x} = 0. 5^{x}

10 0^{\frac{2}{x}} = (1 + 0.1)^{x}

2 (1 0^{x}) + 35 = 179

30 = 286.93 (0.973 8^{x})

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