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

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

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

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

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

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

Solution steps

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

Apply exponent rules

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

Solve

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

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

8^{2 x} - 10 (8^{x}) + 16 = 0

1 6^{x} = \frac{1}{8}

2^{x} \cdot 3^{x} = 216

e^{7 x} = 7 e^{x}

e^{2 x} + e^{x} = 6

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