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 ) )}

4^{3 x} = 32

3^{x + 1} + 3^{x} + 3^{x - 1} = 117

3 (2^{2 x}) = 0.75

5^{x} = 100

6 x = \frac{10}{3} - 6^{- x}

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)))