What is the answer to 3^{2x-1}=5^{x+1} ?

The answer to 3^{2x-1}=5^{x+1} is x=(ln(15))/(2ln(3)-ln(5))

Solve for x: 3^{2x-1}=5^{x+1}

The solution is x=(ln(15))/(2ln(3)-ln(5))

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3^{2 x - 1} = 5^{x + 1}

x = \frac{ln ( 15 )}{2 ln ( 3 ) - ln ( 5 )}

Decimal Notation

x = 4.60719 \dots

Solution steps

3^{2 x - 1} = 5^{x + 1}

Apply exponent rules

(2 x - 1) ln (3) = (x + 1) ln (5)

Solve

(2 x - 1) ln (3) = (x + 1) ln (5) : x = \frac{ln ( 15 )}{2 ln ( 3 ) - ln ( 5 )}

x = \frac{ln ( 15 )}{2 ln ( 3 ) - ln ( 5 )}

9^{x + 3} = 2 7^{x - 1}

- 2 \cdot (- 2)^{n - 1} = (- 2)^{n}

(5^{x - 3}) (3^{2 x - 8}) = 225

2 = a^{x}

x, 2^{x} + 2^{y} = 2^{y} + 4^{x}

What is the answer to 3^{2x-1}=5^{x+1} ?
The answer to 3^{2x-1}=5^{x+1} is x=(ln(15))/(2ln(3)-ln(5))
Solve for x: 3^{2x-1}=5^{x+1}
The solution is x=(ln(15))/(2ln(3)-ln(5))