What is the answer to 7^{x-1}=3^{x+1} ?

The answer to 7^{x-1}=3^{x+1} is x=(ln(21))/(ln(7)-ln(3))

Solve for x: 7^{x-1}=3^{x+1}

The solution is x=(ln(21))/(ln(7)-ln(3))

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

x = \frac{ln ( 21 )}{ln ( 7 ) - ln ( 3 )}

Solution steps

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

Apply exponent rules

(x - 1) ln (7) = (x + 1) ln (3)

Solve

(x - 1) ln (7) = (x + 1) ln (3) : x = \frac{ln ( 21 )}{ln ( 7 ) - ln ( 3 )}

x = \frac{ln ( 21 )}{ln ( 7 ) - ln ( 3 )}

e^{x} - 2 e^{- x} = 0

x, 4^{x} = 5

y, x = y e^{y}

2^{2 x} + 8 (2^{x}) + 7 = 0

4^{x} = 23

What is the answer to 7^{x-1}=3^{x+1} ?
The answer to 7^{x-1}=3^{x+1} is x=(ln(21))/(ln(7)-ln(3))
Solve for x: 7^{x-1}=3^{x+1}
The solution is x=(ln(21))/(ln(7)-ln(3))