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

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

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

solve for

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

x = e^{c} y + 3 e^{c} + 1

Solution steps

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

Apply log rules

x - 1 = e^{c} y + 3 e^{c}

Solve

x - 1 = e^{c} y + 3 e^{c} : x = e^{c} y + 3 e^{c} + 1

x = e^{c} y + 3 e^{c} + 1

lo g_{3} (x - 8) - lo g_{3} (x) = 2

lo g_{x} (10 + 3 x) = 2

\frac{ln ( x )}{x} = 0.0975

lo g_{e} (x) + 4 = lo g_{e} (x + 6)

lo g_{2} (x) - lo g_{2} (16) = 3

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