What is the answer to solvefor x,ln(x^2-y)=0 ?

The answer to solvefor x,ln(x^2-y)=0 is x=sqrt(1+y),x=-sqrt(1+y)

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solvefor x,ln(x^2-y)=0

solve for

x, ln (x^{2} - y) = 0

x = 1 + y, x = - 1 + y

Solution steps

ln (x^{2} - y) = 0

Apply log rules

x^{2} - y = 1

Solve

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

x = 1 + y, x = - 1 + y

2 lo g_{3} (- x + 1) = 6

lo g_{10} (x + 1) = lo g_{10} (1 - x)

lo g_{10} (x) (81) = 4

3 ln (6 x) = - 3

a, lo g_{10} (a^{2} b) = 1

What is the answer to solvefor x,ln(x^2-y)=0 ?
The answer to solvefor x,ln(x^2-y)=0 is x=sqrt(1+y),x=-sqrt(1+y)