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

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

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

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

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

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

Decimal Notation

x = 0.48328 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

9^{x} \cdot 3^{x^{2}} = 8 1^{2}

e^{- x} - 9 e^{- 3 x} = 0

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

1 0^{x} = 3.94

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

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