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/sci/ - Science & Math


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3123164 No.3123164 [Reply] [Original]

since the bases are the same do I write as

(x+6)=1-(x+2)

(x+6)=-x-2


or

(x+6)/log5 = 1 - (x+2)/log5?


solve algebraically? also would my domain be
(-6, + oo )

>> No.3123174

>>3123164
not exactly.

yes, the bases on the 2 logs are the same, but there is that awkward 1 floating out there.

equating the arguments of the log is only valid in this case:

log[b] (A) = log[b] (B)

then, and only then, it follows that A = B.


but since you have

log[b] (A) = 1 + log[b] (B),

you cannot equate the arguments just yet

>> No.3123177

the base of 1 isn't the same, you have to write it as log5(5), so log5(x+6)=log5(5/x+2), so x+6=5/x+2, then just solve

>> No.3123179

>>3123174
mind guiding me through just a little bit? that 1 is making me go FFFUUUUU

>> No.3123189

>>3123179
>>3123179

sure, do as this guy did >>3123177

keeping in mind the identity:
log[b] (X) - log[b] (Y) = log[b] (X/Y)

so, you know that you can rewrite 1 as a log[5]:

1 = log[5] (5)


so your equation becomes:


log[5] (x + 6) = log[5] (5) - log[5] (x + 2)
log[5] (x + 6) = log[5] ( 5 / (x + 2) )

so now we can equate the arguments:

x + 6 = 5 / (x + 2)

and continue on with the algebra!

>> No.3123208

>>3123177
>>3123189

-7 is not E of the Domain?

im thinking the domain is (-6, oo)

>> No.3123230

>>3123208

actually, the domain has to be even more restricted than that.

you're right that -7 isn't in the domain, but is -5?

plugging into the log on the left side, it sure works, but there are still problems with the log on the right side. log[5] (-3) doesn't exist.

so we actually have to choose the narrowest domain so that BOTH logs are real, not just one.

so then the domain has to be (-2, oo)

>> No.3123237

>>3123230
my professor told me to find the logs by looking at the x terms meaning


(x+2) and (x+6) setting them up as inequalities. I was thinking the domain was -6 since it passes through -2

>> No.3123259

>>3123237

okay, now i'm a little confused.
what are we trying to find the domain of?

>> No.3123273

>>3123259
He told us to find the domain by looking at the X terms like i said before so that when i get my 2 result i will know which is the Element of the domain.

>> No.3123279

domain is x>-2 you fucking retard

neither log can be negative

>> No.3123296

>>3123279
geeze calm down im trying to learn this

>> No.3123303 [DELETED] 
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3123303

/sci/

How do you solve log(base2)n = 1x10^8? It's too fucking big, even wolfram alpha can't solve it.

(trying to compute maximum size of problem solvable on a logn order algorithm).

>> No.3123311
File: 40 KB, 578x864, loggraph.png [View same] [iqdb] [saucenao] [google]
3123311

welp....

>> No.3123312

>>3123296
fuck yourself up you faggot ass

best way to do this is

log(x+6) + log(x+2) = 1

log[(x+6)(x+2)] = 1

(x+6)(x+2) = 5

x^2 + 8x + 12 = 5

x^2 + 8x + 7 = 0

(x+1)(x+7)

x = -1 -7, disregard -7 as that gives log(-1) and log(-5) in question

>> No.3123314

>>3123273

yeah okay, i think he wants you to look at the x terms inside the logs.

in our case, that's
x+6 and x+2

then, set up inequalities by knowing the the argument of a logarithm can never be 0 or smaller. so we get:

x+6 > 0, and
x+2 > 0.

which gives us:

x > -6, and
x > -2.

now we always have to chose the more restrictive one. this is because if we chose x>-6, then values like x = -3 are still allowed, but if you plugged those into the log(x+2) function, you'd get the log of a negative number! so, in order for that to not happen, we need to let the domain be x > -2.

was this clear enough? i hope i'm explaining it well

>> No.3123324

>>3123303

n = 2^(10^8) faggot

>> No.3123365

>>3123164

log(x+6) + log(x+2) = 1

log((x+6)*(x+2)) = 1

(x+6)*(x+2) = 5

Use quadratic formula. Win.

>> No.3123372

>>3123365
>Use quadratic formula on simple factorable quadratic

engineer detected

>> No.3123376

>>3123372

> implying i didn't just put that to help the obvious unknowing OP and not because i myself fail at maths

Insensitive prick.

>> No.3123379

>>3123372

almost rolled the fuck of my chair laughing

>> No.3123390

>>3123372

I'd like to see you solve that without the quadratic formula.

>> No.3123400

>>3123395

My BSc in maths says so.

>> No.3123395

>>3123376
sure you are great at math

>> No.3123405

>>3123390
>idiot detected

challenge accepted

(x+6)(x+2) = 5

x^2 + 8x + 12 = 5

x^2 + 8x + 7 = 0

(x+1)(x+7) = 0

x = -1 -7, disregard -7

>> No.3123409

>>3123395

> implying one solves quadratic equations with known coefficients in math.

>> No.3123422

>>3123400
your BSc is in being a faggot.

>> No.3123424

>>3123422

u jelly?

> 300k starting etc.

>> No.3123427

>>3123409
>implying you fail at math and are 12 yrs old

>> No.3123430

>>3123422
>>3123400
>>3123395
>>3123390
>>3123379
>>3123376
>>3123372

This is not /b/. You are the cancer ruining /sci/.

>> No.3123437
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3123437

>>3123430

>> No.3123449

>>3123424
nope

i'm doing phd in math and i'd pretty much forgotten the quadratic formula exists

maybe doing galois theory would have sparked some memory in quadratics being solvable in radicals

>> No.3123453
File: 30 KB, 485x364, notmad1305878703680.jpg [View same] [iqdb] [saucenao] [google]
3123453

>>3123437

It is better to look mad than to look like an idiot underageb&.

Just pointing out the obvious.

>> No.3123458

>>3123430
you are the cancer ruining sci

go to mathforum@drexel or some shit

>> No.3123462

>>3123449

and?

You are pretty much of a social retard if you didn't notice OP is at a level where he needs the quadratic formula to solve quadratic equations.
You working on a PhD doesn't mean everyone else has to.

>> No.3123467

>>3123458

trolls trolling trolls trolling trolls

is it summer yet?

>> No.3123476

>>3123437
ill suck your dick faggot

>> No.3123512

>>3123462
only a fucking idiot teaches quadratic formula prior to teachng factoring. that's like teaching how to use a calculator before teaching integer division.

and if you think it is normal to be learning log equations when you haven't learnt quadratic factoring you are probably doing maths at university of wales bangor or some other home for retards

but you won't ever know how retarded you are because you are a faggot.

>> No.3123532

>>3123512

> only a fucking idiot teaches quadratic formula prior to teachng factoring. that's like teaching how to use a calculator before teaching integer division.

That is exactly what happens today. Do you live under a rock?

> nd if you think it is normal to be learning log equations when you haven't learnt quadratic factoring you are probably doing maths at university of wales bangor or some other home for retards

I actually went to school before studying.

Social retard detected.

umad aspie?

>> No.3123534

look...
1 = log(5)5 right? You have to take 5 on the power 1 to take 5.
so:
log(5)(x+6) = log(5)5 - log(5)(x+2)
and:
log(5)(x+6) = log(5)(5/(x+2))
x+6=5/(x+2)
x^2 + 8x + 12 = 5
x^2 + 8x + 7 = 0
(x + 7)(x + 1) = 0
x = -7 or -1
-7 can't be interpreted, so it's -1

>> No.3123545

>>3123532
>That is exactly what happens today. Do you live under a rock?
maths teacher here (uk)

it isn't in uk

quadratic factorisation is taught prior to quadratic formula and both taught before log equations.

>aspie
aspies don't tend to troll.

>> No.3123547

>>3123532
someone is mad butthurt and jelly

wait. it's you!

>> No.3123556

>>3123545

> aspies don't tend to troll.

He didn't troll, he just showed he was a social retard. Aspie's totally tend to do so.

In america (yes, not just the usa), that is the way it is in most schools.

>> No.3123563
File: 24 KB, 229x300, notmad1305656991769.jpg [View same] [iqdb] [saucenao] [google]
3123563

>>3123547

> implying i am and not you

i quote:

> but you won't ever know how retarded you are because you are a faggot.

obvious sign of butthurt. You should consider leaving this board and maybe all of the internets for good.

>> No.3123566

>>3123545

> implying OP would have known that

>> No.3123575

What the fuck is wrong with you?

Some guy posted a solution and included a standard way of solving quadratic formulas, which works for EVERY quadratic formula, so OP would be able to solve these kind of problems in any case.

What are you all so mad about?

Summerfags all over the place it seems.

>> No.3123584

>>3123556
of course i'm a troll you idiot.

wait: either you think you are a mind reader or think that an adopted internet personality is real.

if the latter then you have some asperger like tendencies yourself.

now post that post priori troll claim pic like a good retard.

>> No.3123589

>>3123563
persist persists persists

hahahah U MAD

now reply with some green text refuting my juvenile nonsense

>> No.3123594

>>3123556
always knew american education sucked ass

>> No.3123596

>>3123584

> implying you haven't been trolled by him not calling you a troll which is reverse trolling i guess?

>> No.3123599

>>3123575
>sage but will still check post

basically economy of effort.

an obviously factorable quadratic takes less work that the formula.

it is therefore a more graceful way to proceed, and grace is at the heart of good mathematics

>> No.3123600

>>3123589

Why are you even posting if you are not mad? It makes no sense.

Therefore, you are mad.

>> No.3123605

>>3123596
so you did do the post priori troll claim, but in reverse and to yourself!

also:samefag

>> No.3123606

>>3123599

> an obviously factorable quadratic takes less work that the formula.
> obvious

you can't tell anyone that somebody who can't solve the equation posted by OP would see at first sight that this is a factorable formula.

troll elsewhere please

>> No.3123608

>>3123600
for fun

aspie detected in green text etc

>> No.3123609

>>3123605

nope

lrn2samefag detect

just joining in on the fun

>> No.3123614

>>3123606
if you can't see x^2 + 8x + 7 is instantly factorable (it doesn't even have negatives, the constant term is prime, the x coefficient is constant + 1. JEEZ SO MANY CLUES) then give up maths now.

>> No.3123617

>>3123609
>samefag claiming not same fag

hahaha trolling u is so very pure and joyful

>> No.3123625

>>3123614

In order to see that you would have to calculate (x+6)(x+2). Why would someone do that for OP?
The post contained a short and general way to solve these problems.

There is NO general way to see if quadratic formulas are (integer) factorable.

>> No.3123630

>>3123606
>so many assumptions

when i was first learning logs i may have stumbled over that unlogged 1, but i would have factored. but then i'm much much better at maths than you.

>> No.3123638

>>3123630

You and I are not OP. So please stop pulling this shit out of your ass. It is not relevant.
If you don't see how it was only to help OP you are the social retard i claimed you to be.

>> No.3123652

>>3123625
>In order to see that you would have to calculate (x+6)(x+2).

you have to do that anyway as quad formula requires quadratic to be equal to zero, so you have to incorporate the 5 on the right hand side

>There is NO general way to see if quadratic formulas are (integer) factorable.

yes there is, if the discriminant is a square number.

like i said, the aspie/troll/social retard in this thread (me) is the only one who can do maths and you are a real genuine maths retard

>> No.3123663

>>3123652

> you have to do that anyway as quad formula requires quadratic to be equal to zero, so you have to incorporate the 5 on the right hand side

Exactly. Why would i do that?

> yes there is, if the discriminant is a square number.

So you are using the quadratic formula to decide whether a quadratic formula is factorable and then factor it? wut?

>> No.3123665

>>3123638
the shit being pulled out of someones ass is yours, by me

yours sincerely

your special troll for the day

>> No.3123686

>>3123663
>Exactly. Why would i do that?
because you have to, whether factoring or using formula to solve, you have to set the thing equal to zero.

if you don't realise that then lol.

>So you are using the quadratic formula to decide whether a quadratic formula is factorable and then factor it? wut?

i wouldn't do it that way, but i am countering your absurd and ignorant suggestion that there is NO (your caps) way to tell if something is factorable in integers. b^2 - 4ac (which is of course only part of the formula) tells you just that.

>> No.3123700

>>3123686

> because you have to

Why would i do that IN HERE, like INSTEAD of OP. Do you even read?

> wouldn't do it that way

What way would you do it?

> i am countering your absurd and ignorant suggestion that there is NO (your caps) way to tell if something is factorable in integers

No way without calculation needed anyway for the quadratic formula.
You say it is more efficient but in order to use it you have to make sure the formula is integer factorable, what is done by calculating the discirminante.

Are you dyslexic? You seem to not get at all what i write. maybe you just ignore me like the 12-year-old you probably are.

>> No.3123721

>>3123700
lol you're still here.

post again. i dare you.

>> No.3123731

>>3123721

ok

>> No.3123738

>>3123731
summerfags everywhere

>> No.3123740

>>3123738

summerfag detected

>> No.3123755

you guys mind helping me on some other problems?

>> No.3123768

Assume the base is log5 but i havent written it

log (x+6)+log (x+2)=1
log x^2+8x+12=1

>> No.3123776

>>3123768

slowpoke tripfaggot

>> No.3123808

>>3123755
bring it on

so long as you don't mind be calling you a retarded piece of shit

my math will be good though

>> No.3123821
File: 10 KB, 688x252, radioactivedecayproblem.png [View same] [iqdb] [saucenao] [google]
3123821

>> No.3123859

2e^(k8) = 1

e^(k8) = 1/2

8k = ln(1/2)

k = ln(1/2)/8

second question just sub in t = 20

suggest you pay attention in class, this is not asking anything beyond the core methods