What is the answer to 7^{2x-1}=5^x ?

The answer to 7^{2x-1}=5^x is x=(ln(7))/(2ln(7)-ln(5))

Solve for x: 7^{2x-1}=5^x

The solution is x=(ln(7))/(2ln(7)-ln(5))

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7^{2 x - 1} = 5^{x}

x = \frac{ln ( 7 )}{2 ln ( 7 ) - ln ( 5 )}

Decimal Notation

x = 0.85257 \dots

Solution steps

7^{2 x - 1} = 5^{x}

Apply exponent rules

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

Solve

(2 x - 1) ln (7) = x ln (5) : x = \frac{ln ( 7 )}{2 ln ( 7 ) - ln ( 5 )}

x = \frac{ln ( 7 )}{2 ln ( 7 ) - ln ( 5 )}

e^{3 t + 6} = 3 e^{6 - t}

25 = 5 (8)^{- 3 x - 1}

7^{x - 1} = 5

1 2^{x^{2} + 2 x + 3} = 1728

2^{2 x} = \frac{1}{8}

What is the answer to 7^{2x-1}=5^x ?
The answer to 7^{2x-1}=5^x is x=(ln(7))/(2ln(7)-ln(5))
Solve for x: 7^{2x-1}=5^x
The solution is x=(ln(7))/(2ln(7)-ln(5))