What is the answer to 3^{x+1}+3^x+3^{x-1}=39 ?

The answer to 3^{x+1}+3^x+3^{x-1}=39 is x=2

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3^{x+1}+3^x+3^{x-1}=39

3^{x + 1} + 3^{x} + 3^{x - 1} = 39

x = 2

Solution steps

3^{x + 1} + 3^{x} + 3^{x - 1} = 39

Factor

3^{x + 1} + 3^{x} + 3^{x - 1} : 13 \cdot 3^{x - 1}

13 \cdot 3^{x - 1} = 39

Divide both sides by

13

3^{x - 1} = 3

Apply exponent rules

x - 1 = 1

Solve

x - 1 = 1 : x = 2

x = 2

e^{- 9 x} = - 3

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

2^{x^{2} - 3 x} = 16

5^{x - 2} - 5^{x} + 5^{x + 1} = 505

4^{\frac{2 z}{3}} = 8^{(z + 2)}

What is the answer to 3^{x+1}+3^x+3^{x-1}=39 ?
The answer to 3^{x+1}+3^x+3^{x-1}=39 is x=2
Solve for x: 3^{x+1}+3^x+3^{x-1}=39
The solution is x=2