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[Jim Groovester]

Oh, that's a really clever.

Anyways, what you need to do is set the equation equal to zero and then solve it as a quadratic. (Make the substitution u = 2^x if you can't see it.) Once you do that you should get stuff that you can do fancy logarithm stuff to. There should be only one solution where you can take the logarithm anyway, and that should be your answer.

I wish I could say I figured this out all on my own but I had to get a little help from WolframAlpha.

[KaminaSquirtle]

Oh, that's a really clever.

Anyways, what you need to do is set the equation equal to zero and then solve it as a quadratic. (Make the substitution u = 2^x if you can't see it.) Once you do that you should get stuff that you can do fancy logarithm stuff to. There should be only one solution where you can take the logarithm anyway, and that should be your answer.

I wish I could say I figured this out all on my own but I had to get a little help from WolframAlpha.

I was about to be all impressed by you thinking of that, but then I read the last line.

[Argembarger]

dat wolfram, man
dat wolfram.

[Fossaman]

Oh, that's a really clever.

Anyways, what you need to do is set the equation equal to zero and then solve it as a quadratic. (Make the substitution u = 2^x if you can't see it.) Once you do that you should get stuff that you can do fancy logarithm stuff to. There should be only one solution where you can take the logarithm anyway, and that should be your answer.

I wish I could say I figured this out all on my own but I had to get a little help from WolframAlpha.

I still don't get it. If I substitute u=2^x and set it equal to zero, I get
2^u - 4u - 21 = 0

...which I can't figure out either.

[KaminaSquirtle]

Oh, that's a really clever.

Anyways, what you need to do is set the equation equal to zero and then solve it as a quadratic. (Make the substitution u = 2^x if you can't see it.) Once you do that you should get stuff that you can do fancy logarithm stuff to. There should be only one solution where you can take the logarithm anyway, and that should be your answer.

I wish I could say I figured this out all on my own but I had to get a little help from WolframAlpha.

I still don't get it. If I substitute u=2^x and set it equal to zero, I get
u² - 4u - 21 = 0

...which I can't figure out either.

Fixed that equation for you.

[Jim Groovester]

I was about to be all impressed by you thinking of that, but then I read the last line.

All of you who were like DUUURRR HOW DO SOLVE IT can shut up.

[Wyrm]

Clever, clever. I ought to remember that trick next time.

Spoiler (click to show/hide)

To solve it now, you complete the square:

u² - 4u + 4 - 4 - 21 = 0
(u-2)² - 25 = 0

solve for u, then you're home free!

u-2 = 5
u=3

2^x = 3

x log 2 = log 3

x = log 3/log 2

[Fossaman]

Okay, that makes a little more sense. You're splitting that up before substituting.

Now I want to ask about this one in class just to see if the instructor does it the same way.

EDIT: Wyrm, that doesn't look right. I think you subtracted where you shouldn't have, I got u=7.

[Vector]

I was about to be all impressed by you thinking of that, but then I read the last line.

All of you who were like DUUURRR HOW DO SOLVE IT can shut up.

Dude, I started doing that.  Then I went "screw that, I'm playing vidjagames now."

[Jim Groovester]

Clever, clever. I ought to remember that trick next time.

Spoiler (click to show/hide)

To solve it now, you complete the square:

u² - 4u + 4 - 4 - 21 = 0
(u-2)² - 25 = 0

solve for u, then you're home free!

u-2 = 5
u=3

2^x = 3

x log 2 = log 3

x = log 3/log 2

I don't this is correct. You should be getting two different solutions, not identical ones.

(u2 - 4u - 21) = 0

(u - 7)(u + 3) = 0

u = 7
u = -3

Since you can't take the negative logarithm:

2^x = 7

And I don't really want to do the logarithm stuff, but it should be easy enough to take away from here. Well, it's not that hard.

x = log7/log2

[Wyrm]

Yeah, yer right. Dangit.

Edit: Fixed—

(u-2)² = 25
u-2 = ±5
u = ±5+2
u = 7, -3

Too often we forget that a positive real has two square roots.

[Christes]

I'm not sure where to post this, but I'll do it here since mathematically-oriented people seem to frequent this thread.  I just saw the most bizarre thing: a real mathematical proof from a real movie!

http://www.youtube.com/watch?v=etbcKWEKnvg

Perhaps it's a bit technical, but it's encouraging to me that someone would actually take the time to research the math when making one of those "mathematical movies" that crop up occasionally.

[KaminaSquirtle]

Okay, those last few posts? Has no one here ever heard of the quadratic equation?

Of course, but there are quite a few cases where factoring is more convenient, like the one above.  In simple to factor cases, the quadratic formula often only introduces needless complexity.

[Vector]

Okay, those last few posts? Has no one here ever heard of the quadratic equation?

Um... yeah, the quadratic equation is needlessly messy in almost all situations one encounters in high school/lower mathematics.

[lordnincompoop]

What, why? It's nice to have a "be all end all" equation, even if it isn't quite there.