Question

Find all real x such that :
lnx+ln(x+1)+ln(x+2)+ln(x+3)=ln(1680)

Answer 1

x = 5

Answer 2

Try exponentiating both sides.\[e^{\ln x + \ln(x+1) + \ln (x+2) + \ln (x+3)}=e^{\ln(1680)}\]

Answer 3

Try using ln(abc) = lna + lnb + lnc
Thus, the left hand side of your equation becomes
ln[x(x+1)(x+2)(x+3)] = ln 1680
Now, you can equate the arguments.
x(x + 1)(x + 2)(x + 3) = 1680
and that should be fairly easy to answer.

Remember that x, x + 1, x + 2, and x + 3 must all be positive numbers.

Answer 4

Clearly 5 and -8 solve the algebraic equation (though -8 doesn't solve the original one). You can divide the fourth degree polynomial by x - 5 and x + 8 and get a quadratic which might have real roots and might have at least one positive real root, so there might be another answer besides x = 5. I'm not going to do it, but you can.