What is the answer to 7^x=4^{2x-1} ?

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

Solve for x: 7^x=4^{2x-1}

The solution is x=(2ln(2))/(4ln(2)-ln(7))

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

x = \frac{2 ln ( 2 )}{4 ln ( 2 ) - ln ( 7 )}

Decimal Notation

x = 1.67694 \dots

Solution steps

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

Apply exponent rules

x ln (7) = (2 x - 1) ln (4)

Solve

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

x = \frac{2 ln ( 2 )}{4 ln ( 2 ) - ln ( 7 )}

(\frac{1}{3})^{x + 3} = 3^{8 x + 5}

5^{x} + 2 5^{x} + 12 5^{x} = - 5

y, x = \frac{e ^{y} - e ^{- y}}{2}

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

e^{3 - 5 x} = 16

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