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

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

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

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

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

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

Decimal Notation

x = - 0.26957 \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 ( 3 ) - ln ( 2 )}{ln ( 2 ) - 2 ln ( 3 )}

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

2^{2 - 3 x} - 8^{x} = - 3

4^{x} - 4^{x - 1} = 6

3^{2 y + 3} = 27

3 6^{2 x - 1} = 6^{3}

4^{x - 1} = (\frac{1}{4})^{2 x}

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