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

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

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solvefor x,ln(y+1)=ln(x-1)+C

solve for

x, ln (y + 1) = ln (x - 1) + C

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

Solution steps

ln (y + 1) = ln (x - 1) + C

Switch sides

ln (x - 1) + C = ln (y + 1)

Move

C

to the right side

ln (x - 1) = ln (y + 1) - C

Apply log rules

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

Solve

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

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

lo g_{5} (3 x^{2}) = lo g_{5} (27)

ln (x) = e^{7}

ln (3 x) - ln (4) = 2

k, lo g_{k} (a) = b - 2

lo g_{x} (3) = 31

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