What is the answer to 7^{1-x}=4^{3x+1} ?

The answer to 7^{1-x}=4^{3x+1} is x=(ln(7)-2ln(2))/(ln(448))

Solve for x: 7^{1-x}=4^{3x+1}

The solution is x=(ln(7)-2ln(2))/(ln(448))

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7^{1 - x} = 4^{3 x + 1}

x = \frac{ln ( 7 ) - 2 ln ( 2 )}{ln ( 448 )}

Decimal Notation

x = 0.09166 \dots

Solution steps

7^{1 - x} = 4^{3 x + 1}

Apply exponent rules

(1 - x) ln (7) = (3 x + 1) ln (4)

Solve

(1 - x) ln (7) = (3 x + 1) ln (4) : x = \frac{ln ( 7 ) - 2 ln ( 2 )}{ln ( 448 )}

x = \frac{ln ( 7 ) - 2 ln ( 2 )}{ln ( 448 )}

e^{x} - 1 = 2 - 2 e^{- x}

2^{- 3 n} \cdot 4^{2 n - 2} = 16

5^{2 x - 7} = 3125

(5^{x}) (5^{x - 1}) = 10

7^{5 x - 12} = 343

What is the answer to 7^{1-x}=4^{3x+1} ?
The answer to 7^{1-x}=4^{3x+1} is x=(ln(7)-2ln(2))/(ln(448))
Solve for x: 7^{1-x}=4^{3x+1}
The solution is x=(ln(7)-2ln(2))/(ln(448))