Hi,
we are just learning about natural logs and I cant seem to figure out a few.. so I was wondering if I could get some help?
Directions: Solve each equation. Use properties of logarithms to simplify each. Use quadratic formula when necessary.
ln (x+1) + ln x = 1
ln 6 + ln x - ln 2 = 3
ln (y +5) + ln (y - 1) = 3
Thanks
1. ln(x+1) would require a taylor series to expand it. However, the left side of the equation is of the form log(x)+log(x+1). (log being the same as ln for right now) Therefore, by the property that says log(ab) = log(a) + log(b), we can rewrite the left side of your original problem as: ln( (x+1) * x ).
Since x(x+1) = x^2 + x , we have: ln(x^2+x) = 1 . Now, take the e of both sides such that we now have: x^2 + x = e^1 = e You now have
x^2 + x - e = 0. Now the dirty, but not difficult work: FACTOR USING QUADRATIC FORMULA: let a = 1 , b =1 , c= -e
Lets make a step easier on us, first. The term inside the the square root in the quadratic formula is called the "discriminant." Lets determine what that is first (don't lose me, this isn't hard). Discriminant = b^2 - 4ac = 1 + 4e
So, now using the quadratic formula: x = (-b +/- sqrt(1+4e)) / 2a
inserting once again our appropriate values for a,b,&c.... we get:
x = (-1 +/- sqrt(1+4e) ) /2
so, we have two results for x:
x = (-1 + (sqrt(1+4e)))/2
x = (-1 - (sqrt(1+4e)))/2
Hold up now, remember we said x must be positive.... the second answer shows x to be negative by virtue of negative minus negative of a positive all over positive... The first answer for x, however, will be positive, since the discriminant is clearly greater than -1. (e = ~2.7ish).
therefore, x=(-1 + (sqrt(1+4e)))/2
2. ln 6 + ln x - ln 2 = 3
ln6 - ln2 = ln(6/2) = ln(3)
ln(3) + ln(x) = ln(3x)
ln(3x) = 3
e^ln(3x) = e^3
3x = e^3
x = (e^3)/3
3. ln (y +5) + ln (y - 1) = 3
the left side is the same as ln( (y+5)(y-1))
so, ln (y^2 +4y -5) = 3
so, we "e" both sides: Remember that e^ln(x) = x
so, y^2 + 4y -5 = e^3
so, y^2 + 4y + (e^3 -5) = 0
you could go about this now with the quadratic formula (perhaps other ways, but stick to this for practice...)
note that the constant when using the quadratic formula is the entire term
(e^3 -5). Thus, c will be (e^3 - 5)
You can complete this problem. If not, post your issues and the wonderful people on this board or I can point you in the right direction.
Good luck.
-Andy
Round to the nearest hundredth. (I'm not quite sure if im doing this right, and im stuck..)
1. ln (5x - 1) - ln x = -2
ln ((5x - 1)/x) = -2
e^-2 = (5x - 1)/x (this is where im stuck)
2. 2 ln e^x + ln e^3x = 4
(divide all by 2)
ln e^x + ln e^3x = 2 (this is where im stuck)
Please and thank-you for your time!
Ashley
1. So we have gotten this far: e^-2 = (5x - 1)/x
I assume you are supposed to solve for x. In that case, lets just do a few things. multiply both sides by x, getting xe^2 = 5x - 1
then subtract 5x from both sides getting xe^2 - 5x = -1
then multiply everything by NEGATIVEone: 5x - xe^2 = 1
Your teach said to evaluated to the nearest 100th, so lets just make e^2
be 7.3890561 (and we will round at the end).
so we have 5x - 7.3890561x = 1
thus, -2.3890561x = 1
and now x = 1/-2.3890561 = -0.418575.....
You are asked to report your answer to the nearest 100th, so lets go with x = -0.42 (note that rounding to the nearest 100th simply means report two digits (rounding the second one if necessary) after the decimal point.
-Andy
[QUOTE=Ashley0903;119931]Round to the nearest hundredth. (I'm not quite sure if im doing this right, and im stuck..)
This second problem is much easier, as you have already done most of the work. Remember that ln(e^x) is simply x
Likewise, another example: ln(e^(-456xyz)) = -456xyz
and check your arithmetic too (as I have the tendency to skip over that as well)
An ln before an e just sends both of those messy terms to hell.
2. 2 ln e^x + ln e^3x = 4
(divide all by 2) ----> ??????
ln e^x + ln e^3x = 2 (this is where im stuck) ---> ????
I would redo your second step (divide all by 2) and go from there.
Hopefully you can finish this one on your own. If you need more direction, don't hesitate to ask again.
-Andy