What is the answer to 2^{x-1}+2^x+2^{x+1}=7 ?

The answer to 2^{x-1}+2^x+2^{x+1}=7 is x=1

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

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

x = 1

Solution steps

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

Factor

2^{x - 1} + 2^{x} + 2^{x + 1} : 7 \cdot 2^{x - 1}

7 \cdot 2^{x - 1} = 7

Divide both sides by

7

2^{x - 1} = 1

Apply exponent rules

(x - 1) ln (2) = ln (1)

Solve

(x - 1) ln (2) = ln (1) : x = 1

x = 1

5 = e^{x}

e^{x} - 5 = 3

e^{x} - 4 = 0

2^{2 x} - 9 \cdot 2^{x} + 8 = 0

3^{x + 1} + 3^{x + 2} = 12

What is the answer to 2^{x-1}+2^x+2^{x+1}=7 ?
The answer to 2^{x-1}+2^x+2^{x+1}=7 is x=1
Solve for x: 2^{x-1}+2^x+2^{x+1}=7
The solution is x=1