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

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

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

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

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

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

Decimal Notation

x = 1.52864 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3^{2 x - 8} = 9^{x - 4}

600 = 4 \cdot 3^{t}

e^{2 x} = 1.4

87 = 1 9^{x}

7^{x + 4} = 4 9^{2 x - 1}

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