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

The answer to 5^{x+1}=7^{-2x-3} is x=-(ln(1715))/(ln(245))

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

The solution is x=-(ln(1715))/(ln(245))

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

x = - \frac{ln ( 1715 )}{ln ( 245 )}

Decimal Notation

x = - 1.35372 \dots

Solution steps

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

Apply exponent rules

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

Solve

(x + 1) ln (5) = (- 2 x - 3) ln (7) : x = - \frac{ln ( 1715 )}{ln ( 245 )}

x = - \frac{ln ( 1715 )}{ln ( 245 )}

2^{x + 1} + 2^{3 x} = 5 \cdot 2^{x + 4}

2^{x + 1} = 8^{x}

x, 2^{2 x} - 2^{x + 1} = 0

x, t (x) = e^{x}

2 5^{x} - 3 \cdot 5^{x} = 0

What is the answer to 5^{x+1}=7^{-2x-3} ?
The answer to 5^{x+1}=7^{-2x-3} is x=-(ln(1715))/(ln(245))
Solve for x: 5^{x+1}=7^{-2x-3}
The solution is x=-(ln(1715))/(ln(245))