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

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

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

solve for

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

x = e^{- c} y - 1

Solution steps

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

Switch sides

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

Move

c

to the right side

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

Apply log rules

1 + x = e^{- c} y

Solve

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

x = e^{- c} y - 1

lo g_{2} (x^{2}) = 6

2 lo g_{4} (x - 1) + lo g_{4} (x - 1) = 6

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

2 ln (x) = ln (6 - x) + ln (3)

lo g_{10} (x) = \frac{1}{2}

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