What is the answer to solvefor x,ln(y-1)=ln(x+3)+c ?

The answer to solvefor x,ln(y-1)=ln(x+3)+c is x=e^{-c}y-e^{-c}-3

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solvefor x,ln(y-1)=ln(x+3)+c

solve for

x, ln (y - 1) = ln (x + 3) + c

x = e^{- c} y - e^{- c} - 3

Solution steps

ln (y - 1) = ln (x + 3) + c

Switch sides

ln (x + 3) + c = ln (y - 1)

Move

c

to the right side

ln (x + 3) = ln (y - 1) - c

Apply log rules

x + 3 = e^{- c} y - e^{- c}

Solve

x + 3 = e^{- c} y - e^{- c} : x = e^{- c} y - e^{- c} - 3

x = e^{- c} y - e^{- c} - 3

1 - ln (x - 1) = 0

lo g_{3} (2 x - 3) + lo g_{3} (5) = 4

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

2 + ln (x + 5) = 7

a (t) lo g_{10} (t) = a (t) y

What is the answer to solvefor x,ln(y-1)=ln(x+3)+c ?
The answer to solvefor x,ln(y-1)=ln(x+3)+c is x=e^{-c}y-e^{-c}-3