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[Johnny Stooge]

I guess should also clarify that I don't aim to be scientiffic at all.

I aim to sound as crazy as possible. All I need is plausible sources.

 

[Dogenzaka]

I got it today. My math tutor came over.

I was doing it completely wrong...I was putting one inside the other and not treating the ln(x) term as something you can't simply throw around and mix with other numbers as if they were like terms.

So basically... -2....ln(5)....x...three different terms, and all he did was write them in a certain order. -2ln(5)x. Gotcha'

It is inaccurately to say k*Log(x)=Log(k*x), instead the correct equivalence is Log(x^k), which is a rather important difference.

THIS

thank you.

Thanks for all the help guys.

I guess my biggest problem now is trying to identify graphs of logarithmic equations and transformations of them, but I'm not sure if you guys can help me with that.

 

[Dogenzaka]

Okay I need help again...the equation is

ln(5) x - ln(2x-1) = ln(4)

I'd be fine with this equation if it wasn't for that x in the ln(2x-1)...I can't extract it. How do I do that? My teacher never showed us how.

I end up with something like ln(8x-4)/ln(5) = x but that's not a good answer :l

 

[Banishing Blade]

Okay I need help again...the equation is

ln(5) x - ln(2x-1) = ln(4)

I'd be fine with this equation if it wasn't for that x in the ln(2x-1)...I can't extract it. How do I do that? My teacher never showed us how.

I end up with something like ln(8x-4)/ln(5) = x but that's not a good answer :l

That equation has no solution (ie, you can't say 'x = constant'). Is there a chance you wrote it down wrong?

 

[The Fishman]

You're thinking right, but as Banishing Blade said, are you sure you typed the problem correctly?

I don't know if it's obvious, but you're right about your answer being wrong because you still have x on both sides, kind of illustrating what B.B. said.

Edit: By the way, thanks for the explanation for that one genetics problem. Sickle-cell isn't sex-linked, so you're right about that. I guess the person was thinking of hemophilia or something.

 

[Dogenzaka]

You're welcome!

And yeah that was the problem written down correctly...I ended up getting the answer but I still don't even know how or anything, I forgot.

Thanks for all your help....I got a 93 on my logarithms test :>

 

[The Fishman]

I got a 93 on my logarithms test :>

Congrats. I know how confusing logs can get sometimes, so good job.

 

[stephaknee]

You're welcome!

And yeah that was the problem written down correctly...I ended up getting the answer but I still don't even know how or anything, I forgot.

Thanks for all your help....I got a 93 on my logarithms test :>

You must have written down your problem wrong-- I have tried everything, checked my TI calculator, numerous online calculators, I have graphed it (I'm a little obsessive when it comes to math problems I can't do :<) and definitely no solution. For proof:

Algebra.Help -- Equation Calculator

Not like it matters anymore! But I've been working on this problem for days >:

To answer your question about how to extract the x in ln(2x-1), though, you would raise everything by e. So it would be e^(ln2x-1) which cancels out to just be 2x-1 (because all ln problems are just log base e).

 

[Dogenzaka]

Stupid long-ass equation.

I'm obsessive with math problems I can't do as well, that's why I've been pissed at this matrix problem that I can't get the right answer to.

"Solve the following linear system using Gaussian elimination. Classify as inconsistent/consistent, independent/dependent:"

2x+y-2z=-1
3x-3y-z=5
x-2y+3z=6

So I turned it into a matrix
2 1 -2 l -1
3 -3 -1 l 5
1 -2 3 l 6

and did row operations and stuff and ended up getting very very odd numbers that weren't the answer.

The answer is:
(-893/180, 17/70, -247/180)

;__;

 

[Banishing Blade]

Those are the wrong answers. The correct answer in (x,y,z) format is (24/13, -3/13, 16/13).

Spoiler Show

2 1 -2 | 1
3 -3 -1 | 5 ~
1 -2 3 | 6

1 -2 3 | 6
3 -3 -1 | 5 swapped R1 and R3 ~
2 -1 2 | 6

1 -2 3 | 6
0 3 -10 | -11 -3*R1 + R2 ~
0 5 -8 | -11 -2*R1 + R3

1 -2 3 | 6
0 1 -12 | -15 -1*R3 +2*R2 ~
0 5 -8 | -11

1 -2 3 | 6
0 1 -12 | -15 ~
0 0 52 | 64 -5*R2 + R3

1 -2 3 | 6
0 1 -12 | -15
0 0 1 | 16/13 1/52 * R3

Solving yields the answer from above.

 

[Dogenzaka]

I KNEW THEY WERE WRONG....my math prof. attempts to thwart me yet again.

Then again my answers were still wrong lol. I got like (-253/36, 15/2, -61/36)

EDIT: Wait I listed the wrong problem :l

It was
7x+33y+z=20
3x+10y+3z=-2
x+3y-5z=7

and the answers are supposed to be (-893/180, 17/10, -247/180) and I got (-253/36, 15/2, -61/36)

That problem I listed above was a different one whose answers were:
(1,-1,1) which is what I got, so I'm not sure how you got your answers of (24/13, -3/13, 16/13). o.o
EDIT: You forgot to change the answer to the first row to 1. You left it as 6.

 

[Hollow Bastion]

I KNEW THEY WERE WRONG....my math prof. attempts to thwart me yet again.

Then again my answers were still wrong lol. I got like (-253/36, 15/2, -61/36)

EDIT: Wait I listed the wrong problem :l

It was
7x+33y+z=20
3x+10y+3z=-2
x+3y-5z=7

and the answers are supposed to be (-893/180, 17/10, -247/180) and I got (-253/36, 15/2, -61/36)

That problem I listed above was a different one whose answers were:
(1,-1,1) which is what I got, so I'm not sure how you got your answers of (24/13, -3/13, 16/13). o.o
EDIT: You forgot to change the answer to the first row to 1. You left it as 6.

Give the correct problem, gosh.

seriously, though,

Spoiler Show

7 33 1 | 20
3 10 3 | -2
1 3 -5 | 7

1 3 -5 | 7
3 10 3 | -2 swap R1 and R3
7 33 1 | 20

1 3 -5 | 7
0 1 18 | -23 R2 = -3R1 + R2
0 12 36 | -29 R3 = -7R1 + R3

1 3 -5 | 7
0 1 18 | -23
0 0 -180 | 247 R3 = -12R2 + R3

1 3 -5 | 7
0 1 18 | -23
0 0 1 | (-247/180) R3 = -1/180R3

1 3 0 | (5/36) R1 = 5R3 + R1
0 1 0 | (17/10) R2 = -18R3 + R2
0 0 1 | (-247/180)

1 0 0 | (-893/180) R1 = -3R2 + R1
0 1 0 | (17/10)
0 0 1 | (-247/180)

 

[Banishing Blade]

It's not wrong, it was just a typo (check the row operations).

(1,-1,1) can't work. Sub those values into the first equation.

 

[Dogenzaka]

Yeah it does...
x =1
y =-1
z =1

2x+y-2z=-1
2(1) + (-1) - 2(1) = -1
2 - 1 - 2 = -1

3x-3y-z=5
3(1) - 3(-1) - (1) = 5
3 + 3 - 1 = 5

x-2y+3z=6
1 -2(-1) + 3(1) = 6
1 + 2 + 3 = 6

They all work.

 

[Banishing Blade]

Oh, I see. I wrote the answer to the first one as positive one, not negative one.

So the answers I gave were right for the matrix I used... >.>

 

[Hollow Bastion]

It's not wrong, it was just a typo (check the row operations).

(1,-1,1) can't work. Sub those values into the first equation.

You can use a graphing calculator to perform rref on a matrix.
And yeah, I about did the same mistake you did. Except with the -2.

 

[Nostalgia]

Trig Help Need:

Cos(2arctan(3/4))

Simple concept, but here's where I'm stuck. The first thing I thought was, "Oh, those are the two legs of a 3,4,5 triangle. This should be no sweat." Then, I realized that knowing the 3 sides won't help with the angle measurements. I'm not supposed to use a calculator, (it's practice for a placement test) and I thought, "Shit. I should memorize the angle measurements for that triangle." which would help with any possible 6,8,10, 12,16,20, and so on. The problem is that I punched tan^(-1)3/4 and it gave me an irrational number. Next, I checked to see if it existed in radical form. It doesn't. So, my question is, how the hell could one hope to solve this problem without a handy calculator? Especially, since that irrational # * 2 is some number that I also have to find the cosine of at the end. I'd really appreciate the help.

 

[Dogenzaka]

I don't know if you guys just like doing matrices or something, but this bitch had me in my seat for quite a while. It wasn't very difficult, just took a long time to finish and lots of room for error :l

Given A:
1 -1 0
3 0 2
-1 0 -1

and A^-1:
0 1 2
-1 1 2
0 -1 -3

are inverses, solve the following system of linear equations without using a calculator:
x-y=4
3x+2z=7
-x-z=3

I got x:
13
9
-16

but man did that take a while

 

[The Fishman]

Trig Help Need:

Cos(2arctan(3/4))

Simple concept, but here's where I'm stuck. The first thing I thought was, "Oh, those are the two legs of a 3,4,5 triangle. This should be no sweat." Then, I realized that knowing the 3 sides won't help with the angle measurements. I'm not supposed to use a calculator, (it's practice for a placement test) and I thought, "Shit. I should memorize the angle measurements for that triangle." which would help with any possible 6,8,10, 12,16,20, and so on. The problem is that I punched tan^(-1)3/4 and it gave me an irrational number. Next, I checked to see if it existed in radical form. It doesn't. So, my question is, how the hell could one hope to solve this problem without a handy calculator? Especially, since that irrational # * 2 is some number that I also have to find the cosine of at the end. I'd really appreciate the help.

Stupid question since most math placement tests aren't, but is this a multiple choice question? Also, I'm guessing you need to find what that exact angle is? This isn't a surefire way to go about doing the problem, but I would try drawing out the unit circle and draw a line with a slope of 3/4 to get an idea of where the point and estimate of the angle is. In correlation to that, you could try drawing a line with slope of 1 since you should know that arctan(1) is equal to pi/4. Since y = (3/4)x is under y = x, you know the angle is a little less than 45 degrees and multiplying it by 2 should still keep you in the first quadrant. So you know your angle should be between about 70 and 90 after multiplying by 2. I know this probably doesn't answer your question or help you at all, but I thought I'd give it a shot and maybe help you see the problem in a different way so you can solve it.

By the way, I believe arctan(3/4) = 36.87 degrees or something like that. So you should have something like cos(73.??). I would guesstimate it to be around .25.

Edit: Or you can try something like these: SOLUTION: Evaluate without a calculator: A) sin{ Arctan [(-5/13)]} B)cos{ Arcsin[(-3/4)]} C)Arccos[(-square root of 3/2)] D) tan { Arcsin[(-3/4)]}
I don't think they would make you simplify these types of problems on a placement exam, but I'm not too sure.

 

[Monkey]

can someone do the ratio test on (3^K)*(K^2) / (K!)