What is the answer to 2^x3^{x-2}=4 ?

The answer to 2^x3^{x-2}=4 is x=(2+2log_{2}(3))/(1+log_{2)(3)}

Solve for x: 2^x3^{x-2}=4

The solution is x=(2+2log_{2}(3))/(1+log_{2)(3)}

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2^{x} 3^{x - 2} = 4

x = \frac{2 + 2 lo g _{2} ( 3 )}{1 + lo g _{2} ( 3 )}

Decimal Notation

x = 2

Solution steps

2^{x} \cdot 3^{x - 2} = 4

Apply exponent rules

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

Solve

x + lo g_{2} (3) (x - 2) = 2 : x = \frac{2 + 2 lo g _{2} ( 3 )}{1 + lo g _{2} ( 3 )}

x = \frac{2 + 2 lo g _{2} ( 3 )}{1 + lo g _{2} ( 3 )}

x^{2} e^{x} - 6 x e^{x} - 7 e^{x} = 0

2 (1 + 0.1275)^{x} = (1 + 0.1375)^{x}

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

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

1 5^{- x + 2} = 1 1^{- 7 x}

What is the answer to 2^x3^{x-2}=4 ?
The answer to 2^x3^{x-2}=4 is x=(2+2log_{2}(3))/(1+log_{2)(3)}
Solve for x: 2^x3^{x-2}=4
The solution is x=(2+2log_{2}(3))/(1+log_{2)(3)}