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

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

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

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

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

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

Decimal Notation

x = 8.63768 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

2^{x} = \frac{1}{32}

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

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

y, e^{y} - p^{3} - p = 0

729 \cdot (2^{n - 5})^{n} - 64 \cdot (3^{n})^{n - 5} = 0

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