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

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

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

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

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

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

Decimal Notation

x = - 2.70951 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

2^{x^{2} - 1} = 0

9^{x + 2} = 27

lo g_{2} (2 5^{x} + 3 - 1) = 2 + lo g_{2} (5^{x} + 3 + 1)

e^{2 x} \cdot (x - 6) = 0

3^{n} + 9 = 81

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