What is the answer to solvefor y,ln(y^2)=x+1 ?

The answer to solvefor y,ln(y^2)=x+1 is y=sqrt(e^{x+1)},y=-sqrt(e^{x+1)}

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solvefor y,ln(y^2)=x+1

solve for

y, ln (y^{2}) = x + 1

y = e^{x + 1}, y = - e^{x + 1}

Solution steps

ln (y^{2}) = x + 1

Apply log rules

y^{2} = e^{x + 1}

Solve

y^{2} = e^{x + 1} : y = e^{x + 1}, y = - e^{x + 1}

y = e^{x + 1}, y = - e^{x + 1}

5.7 = - lo g_{10} (x)

x, z (x) = x ln (1 + 2 x)

2 lo g_{4} (x) - 1 = 0

- lo g_{10} (x) = 0.209

2 ln (x - 3) + ln (x + 7) = 3 ln (x)

What is the answer to solvefor y,ln(y^2)=x+1 ?
The answer to solvefor y,ln(y^2)=x+1 is y=sqrt(e^{x+1)},y=-sqrt(e^{x+1)}