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

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

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

solve for

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

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

Solution steps

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

Apply log rules

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

Solve

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

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

ln (3 x) = - 1

2 lo g_{2} (x) = lo g_{2} (x + 2)

lo g_{10} (x) = 2 - lo g_{10} (x - 15)

x^{l o g_{3} (x)} = 81 x^{3}

lo g_{2} (27) = 3 lo g_{2} (x)

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