What is the answer to 3^{2x-1}=7^{x-1} ?

The answer to 3^{2x-1}=7^{x-1} is x=(-ln(7)+ln(3))/(2ln(3)-ln(7))

Solve for x: 3^{2x-1}=7^{x-1}

The solution is x=(-ln(7)+ln(3))/(2ln(3)-ln(7))

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

x = \frac{- ln ( 7 ) + ln ( 3 )}{2 ln ( 3 ) - ln ( 7 )}

Decimal Notation

x = - 3.37146 \dots

Solution steps

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

Apply exponent rules

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

Solve

(2 x - 1) ln (3) = (x - 1) ln (7) : x = \frac{- ln ( 7 ) + ln ( 3 )}{2 ln ( 3 ) - ln ( 7 )}

x = \frac{- ln ( 7 ) + ln ( 3 )}{2 ln ( 3 ) - ln ( 7 )}

(x - 1)^{(x - 1)} = 27

2^{16 - x} = 2^{x^{2} - 4}

5^{7 x + 5} = 12 5^{2 x - 7}

5^{x + 2} - 5^{x + 3} = 3000

150 = 6 \cdot 5^{x - 1}

What is the answer to 3^{2x-1}=7^{x-1} ?
The answer to 3^{2x-1}=7^{x-1} is x=(-ln(7)+ln(3))/(2ln(3)-ln(7))
Solve for x: 3^{2x-1}=7^{x-1}
The solution is x=(-ln(7)+ln(3))/(2ln(3)-ln(7))