ln(x-4)ln(x+6)= ln e^5 There is no algebraic method for solving this other than by approximation: approximately x = 10.05624144 Did you mean lln(x-4) + ln(x+6)= ln e^5 instead, where they are added instead of multiplied? That can be solved algebraically. You can tell me in the thank-you note form below and I'll get back to you by email. [No charge. I just do this for fun.] Edwin