I need more help with a few math problems, can someone help please? EDIT: 2 more problems i need some help with! :(

[WarDemon666]

EDIT: I need help with 2 more questions, not logs this time, converting.... look down \/



Solve each of the following:

1. ln(x) - ln(x-3) = ln(4)

2. log(x+12) - log(x+3) = 1

3. ln(x) + ln(x+4) = 2 (Hint: Use quadratic formula)


for 1 i got this so far:

ln(x)\(x-3) = ln(4)

didnt know what to do after..

for 2 i got this so far:

log(x+12)\(x+3) = 1

and got stuck again...

working on number 3 right now might get this one....


Thanks for all your help,,, could you guys please post the steps on how to get the answers?

Carl

 

[Oscar1613]

raise both sides to e^(blah)



hint: e^(ln (something)) = something

 

[WarDemon666]

hmmmmm

i hate these problems

 

[WarDemon666]

can anyone give me steps on doing these problems? its the only way ill start to understand

 

[MrScott81]

1)
e^(ln (x/(x-3))) = e^(ln 4)
(x)/(x-3) = 4
x = 4x - 12
3x=12
x=4

 

[WarDemon666]

Originally posted by: scottdog81
1)
e^(ln (x/(x-3))) = e^(ln 4)
(x)/(x-3) = 4
x = 4x - 12
3x=12
x=4

you just canceled out the ln's completly? thats legal?

 

[sash1]

1)
lnx - ln(x-3) = ln4
ln(x/x-3) = ln4
x/x-3 = 4
x = 4(x-3)
x = 4x - 12
3x = 12
x=4

2) same deal

3) x(x+4) = 2
x^2 + 4x - 2 = 0
I'm sure you can take it from there

`K

 

[MrScott81]

yes...like someone mentioned above already:
e^(ln (x)) = x

 

[Jzero]

See Oscar1613's post (e^(ln(x))) == x

 

[MrScott81]

Originally posted by: Kauru

2) same deal

3) x(x+4) = 2
x^2 + 4x - 2 = 0
I'm sure you can take it from there

`K

#3 it should be x^2 + 4x - e^2 = 0 (i think )

 

[TheBoy]

1. x/x-3=4
x=4x-12
-3x=-12
x=4

2.log (x+12)/(x+3)=1 (assuming it's log to the base 10)
(x+12)/(x+3) = 10
x+12 = 10x+30
9x=-18
x=-2

3.ln(x^2 + 4x) = 2
e^2 = x^2 +4x
0 = x^2 + 4x - e^2

I think you can do the formula, i think they're right, don't blame me if theyre not

Edit: In the time i did all that working out someone else went and did it, oh well

 

[WarDemon666]

Originally posted by: scottdog81

Originally posted by: Kauru

2) same deal

3) x(x+4) = 2
x^2 + 4x - 2 = 0
I'm sure you can take it from there

`K

#3 it should be x^2 + 4x - e^2 = 0 (i think )

i tried # 3 and got x^2+4x-2 = 0 and ill just use the quad form to finish? i didnt get e^2...

wheres the problem?

 

[MrScott81]

Originally posted by: WarDemon666

Originally posted by: scottdog81

Originally posted by: Kauru

2) same deal

3) x(x+4) = 2
x^2 + 4x - 2 = 0
I'm sure you can take it from there

`K

#3 it should be x^2 + 4x - e^2 = 0 (i think )

i tried # 3 and got x^2+4x-2 = 0 and ill just use the quad form to finish? i didnt get e^2...

wheres the problem?

because in order to get it to x(x+4) on the left side you have to raise the right side to e^(blah) as mentioned...the same thing as #1, #2

 

[WarDemon666]

how do i factor something like that?!?!

 

[MrScott81]

Originally posted by: WarDemon666

Originally posted by: scottdog81

Originally posted by: Kauru

2) same deal

3) x(x+4) = 2
x^2 + 4x - 2 = 0
I'm sure you can take it from there

`K

#3 it should be x^2 + 4x - e^2 = 0 (i think )

i tried # 3 and got x^2+4x-2 = 0 and ill just use the quad form to finish? i didnt get e^2...

wheres the problem?

because in order to get it to x(x+4) on the left side you have to raise the right side to e^(blah) as mentioned...the same thing as #1, #2

anyways, applying the quadratic formula to x^2+4x-2 you'll get imaginary results and im not sure, but I don't think you can take the ln of an imaginary number???

 

[MrScott81]

sorry double, err, triple post

 

[TheBoy]

Originally posted by: WarDemon666

Originally posted by: scottdog81
1)
e^(ln (x/(x-3))) = e^(ln 4)
(x)/(x-3) = 4
x = 4x - 12
3x=12
x=4

you just canceled out the ln's completly? thats legal?

Yes, very legal, if theyre on both sides

 

[MrScott81]

Originally posted by: WarDemon666
how do i factor something like that?!?!

well e^2 is just a number....
x^2 + 4x - e^2

x = -4 +- sqrt(16- (4)(1)(-e^2)) / 2

x = -4 +- sqrt(16+4e^2) / 2
plugging that into a calculator you get
x is approximately 1.425

you might want to plug it into a calculator to check...i was doing a lot of estimating (especially with my best guess of what e was.

 

[WarDemon666]

Thanks a lot guys, i figured it out. Now im going to go try some more.


You guys are awesome. Thanks again!

 

[WarDemon666]

Im also having trouble with converting stuff (Simplifying?), heres a couple of problems im trying to get:

1. (2x^2 + 7x - 4 / x^4 - 5x^3) / (2x^2 - 32 / x^2 - 5x)

I dont know where to start on this one.... Im starting to get depressed i have to understand all this for an exam in 5 weeks...

2. (6^(-1/3) a^4 b^-3) / ( 6^(5/3) a^-2 b^4 c^3)

edit: I think i have to do soemthing like 6^(-1/3)aaaa-b-b-b / 6^(5/3)-a-abbbbccc????

This one I want to know how to get rid of the negative exponents.....

Thanks again,
Carl

 

[WarDemon666]

^

anyone?

 

[MrScott81]

Originally posted by: WarDemon666
Im also having trouble with converting stuff (Simplifying?), heres a couple of problems im trying to get:

1. (2x^2 + 7x - 4 / x^4 - 5x^3) / (2x^2 - 32 / x^2 - 5x)

I dont know where to start on this one.... Im starting to get depressed i have to understand all this for an exam in 5 weeks...

2. (6^(-1/3) a^4 b^-3) / ( 6^(5/3) a^-2 b^4 c^3)

edit: I think i have to do soemthing like 6^(-1/3)aaaa-b-b-b / 6^(5/3)-a-abbbbccc????

This one I want to know how to get rid of the negative exponents.....

Thanks again,
Carl

don't you have a book?
1. (2x^2 + 7x - 4 / x^4 - 5x^3) / (2x^2 - 32 / x^2 - 5x)
( (2x -1) (x + 4) / (x^2)(x^2-5x) ) / ( (2x-8 ) (x+4) ) / x^2 - 5x)

which simplifies to this:
( (2x - 1) (x+4) (x^2 - 5x) )
-----------------------------------
x^2 ( x^2 - 5x) (2x-8) (x+4)

which simplifies to:
(2x - 1)
-----------------------
(x^2) (2x+8)

I think that's right

 

[Rkonster]

(6^(-1/3) a^4 b^-3) / ( 6^(5/3) a^-2 b^4 c^3)

6^(-1/3) = (6^(1/3))^-1
6^(5/3) = (6^(1/3))^5

 

[WarDemon666]

yeah, i do have a book, but i still dont understand.. lol... :S

i got one of those crap 100 page books with no explantions hardly (or crappy ones)

 

[Rkonster]

Oh, and to get rid of negative exp:

x ^ -a = 1 / x^a