What is the answer to 5^{2x+1}=7^{x-3} ?

The answer to 5^{2x+1}=7^{x-3} is x=-(ln(1715))/(2ln(5)-ln(7))

Solve for x: 5^{2x+1}=7^{x-3}

The solution is x=-(ln(1715))/(2ln(5)-ln(7))

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5^{2 x + 1} = 7^{x - 3}

x = - \frac{ln ( 1715 )}{2 ln ( 5 ) - ln ( 7 )}

Decimal Notation

x = - 5.85025 \dots

Solution steps

5^{2 x + 1} = 7^{x - 3}

Apply exponent rules

(2 x + 1) ln (5) = (x - 3) ln (7)

Solve

(2 x + 1) ln (5) = (x - 3) ln (7) : x = - \frac{ln ( 1715 )}{2 ln ( 5 ) - ln ( 7 )}

x = - \frac{ln ( 1715 )}{2 ln ( 5 ) - ln ( 7 )}

x, y = e^{x} - e^{- x}

e^{2 x} - 3 e^{x} = 0

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

2 0^{x} = 100

2^{a} = 32

What is the answer to 5^{2x+1}=7^{x-3} ?
The answer to 5^{2x+1}=7^{x-3} is x=-(ln(1715))/(2ln(5)-ln(7))
Solve for x: 5^{2x+1}=7^{x-3}
The solution is x=-(ln(1715))/(2ln(5)-ln(7))