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help! i've got a math problem

Submitted by blackbear at 2016-07-02 10:07:54 EDT
Rating: -0.56 on 10 ratings (19 reviews) (Review this item) (V)

i've been searching the internet for fucking chat rooms, math help websites without contact buttons, twitter is just full of morons, so im turning to you fine people, to see if anyone can help me with this.

let's say i've got two numbers, x and y.
if x=2, y=1/2.
if x=4, y=1/4.
if x=10, y=1/10
if x=z, y=1/z
or if x=2^z, y=1/2^z? would that work too?
what is the equation that describes this relationship?

is it x=1/y? if it is, then great. if not, please tell me.

anyways, here's the problem.
x=3, y=1/2
x=9, y=1/4
x=27, y=1/8
x=3^4, y=1/2^4
x=3^z, y=1/2^z
what is that equation? or did i actually just solve it just now? whats going on? help!



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Submitted by ArdAtak at 2017-02-14 16:24:18 EST (#)

Why not just X * Z = 1

Submitted by deathray at 2017-02-14 07:36:10 EST (#)
Rating: 1

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Submitted by Soland at 2017-02-11 21:04:33 EST (#)
Rating: 1

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Submitted by Sabrina Steinman at 2016-12-22 12:06:46 EST (#)
Rating: 2


Submitted by Alex at 2016-12-04 05:13:11 EST (#)


It's Y = 1/X

The last equation is correct but I'd throw in some parenthesis to avoid confustion: Y = 1/(2^Z)

For part 2, it's a logarithm

Z = log3(X) == therefore ==> Y = 1/2^(log3(X))
I know it cause my http://datingbrides.com/blog/dating-russian-girls/first-date-with-a-russian-girl-win-her-heart-from-the-beginning/ is a Math teacher

Submitted by Perry Masterson at 2016-10-07 10:10:23 EDT (#)

God bless I do not have any math problems! Man, use http://writemyessaytoday.net/ and forget about that boring problem solving!

Submitted by blackbear at 2016-07-13 18:44:57 EDT (#)

thanks ardatak, yea i knew it was something simple. i didn't know what logarithms were, but i do now.

Submitted by ArdAtak at 2016-07-12 15:26:59 EDT (#)
Rating: 0

It's Y = 1/X

The last equation is correct but I'd throw in some parenthesis to avoid confustion: Y = 1/(2^Z)

For part 2, it's a logarithm

Z = log3(X) == therefore ==> Y = 1/2^(log3(X))

more about logs: http://math.stackexchange.com/questions/956776/whats-the-inverse-operation-of-exponents

Submitted by Shlongy at 2016-07-11 21:00:06 EDT (#)
Rating: -2

Couch surfer with no 401K, below.

Submitted by OathMeal at 2016-07-06 21:49:08 EDT (#)

Fags below.

Submitted by Shlongy at 2016-07-04 10:08:59 EDT (#)
Rating: -2

I was told there'd be no math...

Submitted by HurtByTheSun at 2016-07-04 07:49:13 EDT (#)
Rating: -2

Die.

Submitted by BLITZKREIG_BOB at 2016-07-03 00:35:36 EDT (#)
Rating: -2

Are you supposed to plot your answers in a Cartesian grid?

Either way, do your own fucking homework.

Submitted by blackbear at 2016-07-02 18:26:31 EDT (#)

alright, i think i got this. lets see.
x=2, y=1/2
x=4, y=1/4
x=z, y=1/z

obviously x=1/y works, but also
log2(x)=log.5(y) works too, no?

plug in 4 for x, i get

log2(4) = 2 = log.5(y)

so now i have to undo the log.5(y) to solve for y. what that equation means is that .5 squared equals y. .5 squared, or 1/2 squared is 1/4. y=1/4. it works! sweet!

so now for
x=3, y=1/2
x=9, y=1/4
x=27 y=1/8

we will use log3(x) = log.5(y)
plug in 3^9 for x, we have log3(3^9) = log.5(y)
log3(3^9) = 9 = log.5(y)
so .5^9 = y = 1/2^9 = 1/512.
x= 3^9 = 19683
y = 1/2^9 = 1/512
alright, so now for the ultimate test. i will plug in a number for x i don't already know what the answer is, and see if i get the right result.
x = 17, log3(17) = log.5(y)
solving for y.
log3(17) = 2.578 = log.5(y)
so that means .5 to the 2.578 power equals y.
1/2 ^ 2.578 = .167

so if that's correct, then we have
x=3, y=1/2
x=9, y=1/4
x=27, y=1/8
x=17, y=.167

i think i've solved my problem. excellent!

thanks for all your help everyone. goodnight.

Submitted by Bestmate2 at 2016-07-02 17:01:15 EDT (#)
Rating: -1

Original granted but so cucking boring. X

Submitted by blackbear at 2016-07-02 13:33:34 EDT (#)

i NEED to solve this.

Submitted by blackbear at 2016-07-02 13:31:33 EDT (#)

thanks for that website, tormentos. i read a little about logarithms, which seem to be the opposite of exponents, like multiplication is to division, or addition is to subtraction.
but i still don't know the damn equation!
x=3 y=1/2
x=9 y=1/4
x=27 y=1/8
all im doing is multiplying x by 3, y by 1/2 each time. what equation is that? it should be very similar to x=1/y, but with one side multiplied by something, or to a certain power, or maybe some logarithmic action, i dont know. shit.

Submitted by blackbear at 2016-07-02 13:13:06 EDT (#)

yea silverwolf, that is horrible. it's not x=4y or x=6y, that would be an incredibly simple straight line, and it only works for the first line. the next one down, x=4, y=1/4, 4y equals 1, which is not equal to 4. down below x=9, y=1/4, 6y is 6/4, which is not equal to 9. total fail. but thank you for trying.

Submitted by Tormentos at 2016-07-02 12:57:29 EDT (#)
Rating: -2



Mmm...incapacitating.

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