What is the answer to 3^{x+2}+3^{x+1}+3^x+3^{x-1}=120 ?

The answer to 3^{x+2}+3^{x+1}+3^x+3^{x-1}=120 is x=2

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3^{x+2}+3^{x+1}+3^x+3^{x-1}=120

3^{x + 2} + 3^{x + 1} + 3^{x} + 3^{x - 1} = 120

x = 2

Solution steps

3^{x + 2} + 3^{x + 1} + 3^{x} + 3^{x - 1} = 120

Factor

3^{x + 2} + 3^{x + 1} + 3^{x} + 3^{x - 1} : 40 \cdot 3^{x - 1}

40 \cdot 3^{x - 1} = 120

Divide both sides by

40

3^{x - 1} = 3

Apply exponent rules

x - 1 = 1

Solve

x - 1 = 1 : x = 2

x = 2

9^{x} - 7 \times 3^{x} - 18 = 0

\frac{8 1 ^{3 n + 2}}{24 3 ^{- n}} = 3^{4}

12 e^{2 x} = 200

2^{x} \times 3^{x} = 81

- x^{2} e^{- x} + 2 x e^{- x} = 0

What is the answer to 3^{x+2}+3^{x+1}+3^x+3^{x-1}=120 ?
The answer to 3^{x+2}+3^{x+1}+3^x+3^{x-1}=120 is x=2
Solve for x: 3^{x+2}+3^{x+1}+3^x+3^{x-1}=120
The solution is x=2