What is the answer to 4=sqrt(2e^0+c_{1)} ?

The answer to 4=sqrt(2e^0+c_{1)} is c_{1}=14

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4=sqrt(2e^0+c_{1)}

4 = 2 e^{0} + c_{1}

c_{1} = 14

Solution steps

4 = 2 e^{0} + c_{1}

Square both sides:

16 = 2 + c_{1}

16 = 2 + c_{1}

Solve

16 = 2 + c_{1} : c_{1} = 14

c_{1} = 14

Verify Solutions:

c_{1} = 14

True

The solution is

c_{1} = 14

- x + 1 + 2 = 0

16 x^{2} = 4 x

9^{2} + 8^{2} + 3^{2} = x + 5

3 x + 4 = x + 18

44 - 8 x = x - 7

What is the answer to 4=sqrt(2e^0+c_{1)} ?
The answer to 4=sqrt(2e^0+c_{1)} is c_{1}=14
Solve for e: 4=sqrt(2e^0+c_{1)}
The solution is c_{1}=14