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

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

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

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

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

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

Decimal Notation

x = - 1.19126 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

\frac{5 ^{2 x} - 1}{5 ^{x} + 1} = 4

8^{1 - 2 x} = 16

ln (e^{2 x}) = 2 x

4 e^{3 x} - 7 = 1

(e^{4})^{3 x} = e^{4} e^{3 x}

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