# Thread: Help with natural logarithms?

1. ## Help with natural logarithms?

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

2. Originally Posted by Ashley0903
ln (x+1) + ln x = 1
These all use the same formula to simplify the logarithms:
$log(a) + log(b) = log(ab)$

$log(a) - log(b) = log \left ( \frac{a}{b} \right )$
(where the logs are to any base.)

So for this one:
$ln(x + 1) + ln(x) = 1$

$ln((x + 1)x) = 1$

$(x + 1)x = e$

and I'm sure you can finish this from here?

-Dan

3. actually, thats where im lost, i understand the properties... but after that im confused

4. It tells us to use the quadratic formula, so why not,
$(x+1)x=e\ \ \Rightarrow\ \ x^2+x-e=0$
So
$x=\frac{-1\pm\sqrt{1+4e}}{2}$
But since we require $x>0$, there is an extraneous solution. So the only answer is
$x=\frac{-1+\sqrt{1+4e}}{2}$

5. thats what i thought was supposed to be done, we are also required to have our answers in decimal form rounded to the nearest hundredth an also since you cant have a negative log it would then be 1.22?

6. im still not quite sure.. But all the same, thanks for all the help!

7. Originally Posted by Ashley0903
im still not quite sure.. But all the same, thanks for all the help!

8. the logarithm above that im trying to work out. ive just found this site and really not sure how things are supposed to be done, im just thankful for all the help.

9. ## Here's some help

Originally Posted by Ashley0903
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 (take note NOW that x must be positive, since ln(x) when x less than or equal to zero is invalid. x must be positive!)

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

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

10. Thank you so much Andy, i understood all of it except the last bit of number three... the book says that the answer is 3.393 but i am not coming up with that
I am so grateful for your time
Ashley

11. nevermind, i figured out the problem. For number three, instead of ending up with y^2 + 4y + (e^3 -5) = 0 it should have been y^2 + 4y + (-e^3 -5) = 0. I do thank you oh so very very much for helping me.

12. ## a bit more help on natural logarithms please :)

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)

Ashley

13. Originally Posted by Ashley0903
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)

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

14. Originally Posted by abender
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(what happened to the negative?) = 5x - 1
then subtract 5x from both sides getting xe^2 - 5x = -1
then multiply everything by one: 5x - xe^2 = 1(why cant we just add one to each side?
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
..

15. ## second prob, you've almost done it already

[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

Page 1 of 2 12 Last