What is the answer to 5^{x+1}=7^{x-1} ?

The answer to 5^{x+1}=7^{x-1} is x=(ln(35))/(ln(7)-ln(5))

Solve for x: 5^{x+1}=7^{x-1}

The solution is x=(ln(35))/(ln(7)-ln(5))

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

x = \frac{ln ( 35 )}{ln ( 7 ) - ln ( 5 )}

Solution steps

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

Apply exponent rules

(x + 1) ln (5) = (x - 1) ln (7)

Solve

(x + 1) ln (5) = (x - 1) ln (7) : x = \frac{ln ( 35 )}{ln ( 7 ) - ln ( 5 )}

x = \frac{ln ( 35 )}{ln ( 7 ) - ln ( 5 )}

9^{x} - 4 \cdot 3^{x + 1} + 27 = 0

5^{x^{2} - 12} = 2 5^{2 x}

2 x^{- 3} e^{5 l n (x)} - x - 3 = 0

2^{x + 2} - 2^{x} = 96

5^{x - 3} = 1

What is the answer to 5^{x+1}=7^{x-1} ?
The answer to 5^{x+1}=7^{x-1} is x=(ln(35))/(ln(7)-ln(5))
Solve for x: 5^{x+1}=7^{x-1}
The solution is x=(ln(35))/(ln(7)-ln(5))