I've been working on this natural logarithm problem for a couple days now and I can't figure how to get it to work. Everytime I check my answers it doesn't work. I'll show you what I did so you can show me what I'm doing wrong.
ln(x^2-6x-16) = 5
x^2-6x-16 = e^5
(x-8)(x+2) = e^5
x-8 = e^5
x = e^5 + 8
x+2 = e^5
x = e^5 - 2
{e^5+8,e^5-2}
These answers do not work though. I'd appreciate any help.
2007-02-06 06:37:57 · 4 answers · asked by Anonymous in Education & Reference ➔ Homework Help
When I say the answers don't work, I mean that when I plug them back in they do not satisfy the equation. Should I try substitution because that doesn't seem like it would help, or should I factor it before I eliminate the natural log? I'm really unsure of where I'm going wrong.
2007-02-06 13:19:23 · update #1
ln(x^2-6x-16) = 5
x^2-6x-16 = e^5
You can't factor and set each equation = to e^5, otherwise, when you multiplied them together, you'd get e^10. Instead, use the quadratic formula:
x^2-6x-16 = e^5
x^2 - 6x - (16 + e^5) = 0
x = (-b +/- (b^2 - 4ac)^(1/2)) / 2a
x = (6 +/- (36 - 4(1)(-(16 +e^5)))^1/2) / 2
x = 3 +/- ((693.65)^(1/2) / 2)
x = 3 +/- 13.169
x = 16.169, -10.169 (solution!)
2007-02-08 07:10:35 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
You went wrong when you set the two factors equal to e^5. If two numbers multiply to 0, one must be 0. If two numbers multiply to e^5, one does not have to =e^5.
You should always set quadratic equations = to 0. If you do that here, you get x^2-6x-16-e^5=0. Use the Quadratic Formula with a=1, b=-6 and c=-18-e^5
2007-02-10 11:01:22 · answer #2 · answered by King 2 · 0⤊ 0⤋
It's been a while since I've done this type of math, but from what I remember, it seems like you did the problem right. Do you know what the answers are suppose to be? When you say the answers don't work, what don't they work to? Maybe answering these questions will help people try to solve the problem. Good luck.
2007-02-06 12:15:50 · answer #3 · answered by Alley 2 · 0⤊ 0⤋
ln a million/(a million + e^-x) to ln a million/(e^x + a million)/e^x how did they make it constructive? Is there a rule or something? that is how e^(-x) = a million / e^x. use this formulation you receives it. If nevertheless you do not realize then examine "SURD" in math. O.ok.
2016-11-02 12:20:08 · answer #4 · answered by bason 4 · 0⤊ 0⤋