What is the answer to solvefor x,y^2=e^{2x}+c ?

The answer to solvefor x,y^2=e^{2x}+c is x=(ln(y^2-c))/2

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solvefor x,y^2=e^{2x}+c

solve for

x, y^{2} = e^{2 x} + c

x = \frac{ln ( y ^{2} - c )}{2}

Solution steps

y^{2} = e^{2 x} + c

Switch sides

e^{2 x} + c = y^{2}

Move

c

to the right side

e^{2 x} = y^{2} - c

Apply exponent rules

2 x = ln (y^{2} - c)

Solve

2 x = ln (y^{2} - c) : x = \frac{ln ( y ^{2} - c )}{2}

x = \frac{ln ( y ^{2} - c )}{2}

5^{x} = 18

m, x^{m} = 0

2^{2 x} - 10 \cdot 2^{x} + 16 = 0

8^{x + 1} = 1 6^{x}

e^{2 x} - 7 e^{x} + 12 = 0

What is the answer to solvefor x,y^2=e^{2x}+c ?
The answer to solvefor x,y^2=e^{2x}+c is x=(ln(y^2-c))/2