What is the answer to 2(3^{(2x-1)})=4^{(x+1)} ?

The answer to 2(3^{(2x-1)})=4^{(x+1)} is x=(ln(6))/(2(ln(3)-ln(2)))

Solve for x: 2(3^{(2x-1)})=4^{(x+1)}

The solution is x=(ln(6))/(2(ln(3)-ln(2)))

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2 (3^{(2 x - 1)}) = 4^{(x + 1)}

x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

Decimal Notation

x = 2.20951 \dots

Solution steps

2 (3^{(2 x - 1)}) = 4^{(x + 1)}

Apply exponent rules

ln (2) + (2 x - 1) ln (3) = (x + 1) ln (4)

Solve

ln (2) + (2 x - 1) ln (3) = (x + 1) ln (4) : x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

5^{x^{2}} = 5

1 0^{2 x - 3} = 1000

2 e^{2 x} + 15 = 13 e^{x}

2^{2 x} = 500

x, 3^{4 x + 1} = 27

What is the answer to 2(3^{(2x-1)})=4^{(x+1)} ?
The answer to 2(3^{(2x-1)})=4^{(x+1)} is x=(ln(6))/(2(ln(3)-ln(2)))
Solve for x: 2(3^{(2x-1)})=4^{(x+1)}
The solution is x=(ln(6))/(2(ln(3)-ln(2)))