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

The answer to 7^{1-x}=3^{2x+1} is x=(ln(7)-ln(3))/(ln(63))

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

The solution is x=(ln(7)-ln(3))/(ln(63))

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

x = \frac{ln ( 7 ) - ln ( 3 )}{ln ( 63 )}

Decimal Notation

x = 0.20450 \dots

Solution steps

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

Apply exponent rules

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

Solve

(1 - x) ln (7) = (2 x + 1) ln (3) : x = \frac{ln ( 7 ) - ln ( 3 )}{ln ( 63 )}

x = \frac{ln ( 7 ) - ln ( 3 )}{ln ( 63 )}

(2)^{x + 2} = (0.5)^{2 x - 4}

100 = e^{- 9 t + 3 l n (2)}

5^{x} = e^{x - 1}

x^{l o g_{5} (x) - 2} = 125

5^{4 X + 20} = 5^{12 X + 20}

What is the answer to 7^{1-x}=3^{2x+1} ?
The answer to 7^{1-x}=3^{2x+1} is x=(ln(7)-ln(3))/(ln(63))
Solve for x: 7^{1-x}=3^{2x+1}
The solution is x=(ln(7)-ln(3))/(ln(63))