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Tryska
02-01-2002, 09:41 AM
we are having a major issue at work and it has us all stumped..cuz we are tech geeks....we don't deal with this stuff.

anyhoo....

2x(that's supposed to be the power of x)-2=2

ie: 2 to the x power - 2 = 2

we need to know the steps to derive x. (we already know the answer is 2)

where everyone is getting stumped is as follows:

2 to the x power - 2=2
2 to the x power - 2+2=2+2
2to the xpower = 4
**now what??**
x=2

:help:

Spiderman
02-01-2002, 10:09 AM
Hmmm... I think you then take the square root of each side to get rid of the 2to the xpower = 4. You then will have x = 2. I think thats how it is... :scratch:

Sinep
02-01-2002, 10:12 AM
to bring down the power you have to use log's if I remember right

Tryska
02-01-2002, 10:13 AM
right...that's what i thought to.

BUT.....if we did that, aren't we then assuming that x=2?

you know what i mean?

if x really = 3 instead, taking the square of it, wouldn't really work right?

Tryska
02-01-2002, 10:14 AM
Originally posted by Sinep
to bring down the power you have to use log's if I remember right

ok...so how would you do that?

Podium Kreatin
02-01-2002, 10:16 AM
err.. it's hard to understand from the way it is typed, i think it'll be easier to do if u can scan it and attach it.

Tryska
02-01-2002, 10:18 AM
here ya go.....

02-01-2002, 10:25 AM
2x - 2 = 2

2 x = 4

square root of both sides:

/2x = 2

x = 2

Why is HTML switched off clihgie you wan ker. tuttut

Tryska
02-01-2002, 10:33 AM
so chigs, you are saying:

02-01-2002, 10:42 AM
:nod:

2 ^2 is the same as 4

/4 is the same as 2.

Tryska
02-01-2002, 10:47 AM
but again...by squaring each side, aren't we assuming that x=2?

i mean..what if it was really 2 to the 3rd power, and not the 2nd?

02-01-2002, 10:51 AM
we're not squaring each side - we're taking the square ROOT of each side.

02-01-2002, 10:51 AM
oh btw, if it was 2^3 then 2^x -2 = 2 would be false.

Tryska
02-01-2002, 10:51 AM
that's what i meant.

either way...we are still assuming we're dealing with the 2nd power, by taking the square root, no?

Tryska
02-01-2002, 10:52 AM
bah,

okay..solve this one smart guy:

2^x - 2 = 254

Tryska
02-01-2002, 10:53 AM
and don't give me the answer without showing me how you arrived at the answer.

big
02-01-2002, 10:59 AM
:swear: 2^(x - 2) = 254 ????
or (2^x) - 2 = 254

PowerManDL
02-01-2002, 11:01 AM
Originally posted by Tryska
bah,

okay..solve this one smart guy:

2^x - 2 = 254

x = 16

02-01-2002, 11:02 AM
with that equation, Tina, you can't just take square roots.

If you think of squaring and taking the square root of as being exact opposites.[a bit like + and -]

so you can;t just take the square root of 256 [/256] and expect to get 8 [the value of x in this case.]

02-01-2002, 11:02 AM
Matt you tit, 2^16 = 65536.

PowerManDL
02-01-2002, 11:03 AM
Originally posted by PowerManDL

x = 16

make that x = 8

Duh.

Tryska
02-01-2002, 11:03 AM
wrong. x=8.

big:

(2^x)-2=254

Tryska
02-01-2002, 11:04 AM
with that equation, Tina, you can't just take square roots.

If you think of squaring and taking the square root of as being exact opposites.[a bit like + and -]

so you can;t just take the square root of 256 [/256] and expect to get 8 [the value of x in this case.]

my point exactly.

so again..my question is NOT what x=, but

HOW step by step do i derive what x=??? (regardless of what x might equal)

heathj
02-01-2002, 11:32 AM
2^x - 2 = 2

2^x = 4

2^x^(1/x) = 4^(1/x)

2 = 4^(1/x)

Now....4 to the 1/what power, makes 4 equal two. It has to be 2, because 4^(1/2) = 2. 4^(1/2) is the same thing as the square root of four..Get what I'm saying or is that basically the same as everyone else. You also have to raise 2^x to the (1/x) power to get rid of the power on the left side.

Tryska
02-01-2002, 11:36 AM
okay. that is more right than anything else i've seen..and actually someone did come up with that, but we weren't sure that was right.

can i get verification from one of you math geniuses please?

not that i doubt you heathie..thanks! :thumbup:

Tryska
02-01-2002, 12:24 PM
okay.....we just tried that..and that didn't work either:

(2^x)-2 = 254
2^x -2+2= 254+2
2^x=256
2^x*1/x=256*1/x
2=256/x
2*x=256/x*x
2x=256
2x*1/2 = 256*1/2
x=256/2
x=128

which is not the same as x=8. :rolleyes:

damn subnetting shite pissing me off, no real life fricken value....:mad:

IceRgrrl
02-01-2002, 12:26 PM
Sinep was right...you need to use logarithms (log).

2^x - 2 = 4

Adding 2 to both sides gives you:
2^x = 4

Take the log of both sides:
log (2^x) = log(4)

Now log (2^x) is the same thing as x log (2), so:
x log(2) = log(4)

Divide both sides by log(2) to isolate x:
x =[log(4)]/[log(2)]

x = 2

So by that same process:
2^x - 2 = 254
2^x = 256
log(2^x) = log (256)
x log( 2) = log( 256)
x = log(256) /log (2)
x = 8

Tryska
02-01-2002, 12:28 PM
you are ace! :bow:

ps - my korean co-worker, Sum Dum Guy, says thank you! (he told me to write that *lol*)

IceRgrrl
02-01-2002, 12:30 PM
No problem...glad to be of help. :)

Tryska
02-01-2002, 12:32 PM
bah.

after really looking through it, he's changed his mind.

he says he knows how to do it useing log, but he has to figure out how to do it without a calculator. :confused:

IceRgrrl
02-01-2002, 12:37 PM
Without a calculator? What is this? The Jurassic Age? ;)

log(x) = the power that 10 must be raised to to get x

log (2) = .301029995, so 10^0.3010299957 = 2

That might be a little tough by hand....

Tryska
02-01-2002, 12:56 PM
*lol*

yeah...i'd say. he seems to think there's some sort of way he can do it withough using log. I think he's smoking crack.

big
02-01-2002, 02:28 PM
What does he need it for?

Tryska
02-01-2002, 02:30 PM
he's taking a Network Engineering exam and apparently they can't use calculators or some such thing.

he would need that in the event he needed to figure out subnet masking for 2 particular IP addresses.

Mystic Eric
02-01-2002, 02:36 PM
Yes, this is basic algebra help, and yes, there is a way to do it without logs: when you have a variable (in this case, x) as an exponent, you can make both the bases on each side equal the same. In this question, you would make the bases both equal 2 because you always make the base the smaller number out of the two. Here is the example:

(2^x) -2 =2
(2^x) -2 +2 = 2 + 2 <----- add two to both sides
2^x = 4
2^x = 2^2 <----- using the same base rule making both bases equal 2
therefore:
x=2

And for the next one:

2^x - 2 = 254
2^x = 256
now make both bases the same:
2^x = 2^8
therefore
x = 8

You can do this 3 ways:
1. logs
2. same base rule
3. keep on plugging into your calculator through trial and error. although option number 3 is kinda childish.

and heath, your way is wrong. try doing that method when you get and equation like:
log2 16^(2x+1) = 8

heathj
02-01-2002, 02:39 PM
That equation didn't have logs in the first place did it? Plus with that 254 one..it would've ended up being 2 = 256 ^ (1/x)

And 256 ^ (1/8) = 2....but who cares, math sucks. Sorry Term, thought you weren't posting here anymore.

Mystic Eric
02-01-2002, 02:41 PM
Heath J, you are a dork. The equation didn't have logs in the first place, but you ADD logs to solve equations. Or an other example, you ADD numbers when you are completing the square, but you also subtract the numbers you add so it equals 0 again. Go back to school, junior.

the_hall
02-01-2002, 02:45 PM
That was an easy problem, i figured it out before I even looked at the responces.

Where do you work, the tampon factory?

heathj
02-01-2002, 02:46 PM
Sorry I'm not as smart as you.

Tryska
02-01-2002, 04:07 PM
the-hall....the answer wasn't the tough part assface, deriving the answer using algebraic steps was.

eric....dude..you are cool with me.....it definitely makes perfect sense to multiply the 2 on both sides. great work. sum dum guy is gonna be very pleased. dunno why none of use thought of that. (prolly cuz i'm the youngest, and I have done algebra in over 10 years, and the rest of them are way older than me). thanks for helping!

:thumbup:

the_hall
02-01-2002, 04:14 PM
Originally posted by Tryska
the-hall....the answer wasn't the tough part assface, deriving the answer using algebraic steps was.

I knew the method too!

Tryska
02-01-2002, 04:22 PM
so then why didn't ya post it?

the_hall
02-01-2002, 04:25 PM

Tryska
02-01-2002, 04:26 PM
likely story....

the_hall
02-01-2002, 04:28 PM
Originally posted by Tryska
likely story....

You underestimiate my intelligence. Mathmatics is my game. Im a walking calculator, a TI-92.

Tryska
02-01-2002, 04:34 PM
no..i don't necessarily underestimate it.

but since your such a smarty pants....explain this to me....

2^x - 2 = 254
2^x = 256
now make both bases the same:
2^x = 2^8
therefore
x = 8

how come he went from:

2^x=256 to 2^x = 2^8 ??

i don't get it. i mean obviously he just "did the math" as it were, but is that the next logical step? if so, why?

Podium Kreatin
02-01-2002, 04:47 PM
here's the work (using alg knowledge) (i could've solved it earlier, but i had school, solli :))

Tryska
02-01-2002, 04:51 PM
well thanks man...i appreciate it, but without the logs..do you concur with eric on the same base rule?

the_hall
02-01-2002, 04:52 PM
I don't see what your saying?

All he did was change 256 to 2^8. It is the same number, written differently.

Mystic Eric
02-01-2002, 04:52 PM
Originally posted by Tryska

2^x - 2 = 254
2^x = 256
now make both bases the same:
2^x = 2^8
therefore
x = 8

how come he went from:

2^x=256 to 2^x = 2^8 ??

i don't get it. i mean obviously he just "did the math" as it were, but is that the next logical step? if so, why?

tryska:

that step:
"2^x=256 to 2^x = 2^8"

is how you would make the bases the same. since the base of the left side is 2, you would make the base of the right side 2 as well. The reason behind this is that if the bases are the same, then the exponents on both sides would equal each other so you can just simply solve for x. "2^8" is 256 but you're just making it 2^8 to make the base the same as the left side for the reasons i stated earlier. Hope this helps.

Tryska
02-01-2002, 04:57 PM
okay....but i guess my question is....

in the proper method for an algebraic proof, you can make this conversion (256 --->2^8) without having points taken for not fully showing how you got there?

you know what i mean? or am i beign confusing?

i mean to me..yes that's logical...because i know it's the correct answer, but in algebra is that also considered correct? or considered an assumption?

(i never was good in algebra, because i could arrive at the answer, but not proof my way there)

Mystic Eric
02-01-2002, 05:05 PM
uhhh podium, you normally wouldn't use ln (the natural log) unless you are dealing with expressions that include "e" - the constant that is 2.71828.

tryska, if the bases are the same, then the exponents are equal to each other. Here is a question directly from my "Theory and Problems for Senior High Math" book that will help illustrate this further:

solve for x:

3^(x^2)= 9 * (3^-x) the question would read: Three to the x squared is equal to 9 times 3 to the negative x
to equate this you would do the following,

3^(x^2)= ((3^2) * (3^-x)) <----- all the bases are made to be 3
according to exponential rules you would add the exponents on the same side:

3^(x^2)= 3^(-x+2)

Now, you see how the base on both sides is 3; because of this, the exponents are equal to each other.

x^2 = -x+2 < ---- move everything to one side:

x^2 +x - 2 = 0 <----- now factor

(x+2)(x-1) = 0

x = -2, 1

so all you have to do is either take the logs of both sides, or use the same base rule to solve for x.

Tryska
02-01-2002, 05:08 PM
ace.

thanks for popping up and explaining eric. :)

Mystic Eric
02-01-2002, 05:09 PM
Originally posted by Tryska
okay....but i guess my question is....

in the proper method for an algebraic proof, you can make this conversion (256 --->2^8) without having points taken for not fully showing how you got there?

you know what i mean? or am i beign confusing?

i mean to me..yes that's logical...because i know it's the correct answer, but in algebra is that also considered correct? or considered an assumption?

(i never was good in algebra, because i could arrive at the answer, but not proof my way there)

tryska, using same base rule formula would get you full marks in any math exam. However, you can only use this if the numbers are small like with your questions. If the numbers are larger, if they cannot be made into the same base easily, or if the equation is complex, then one MUST use logs. the example i showed earlier and the way to do it was directly from my math book and this guy makes the provincial examinations in 3 provinces in canada. tryska, trust me, i would not mislead you, would i? ;)

Tryska
02-01-2002, 05:11 PM
well i'm trusting you on this one eric....since it makes sense to me..and your the only person that popped up with it. i appreciate it...and i will pass it on to my co-worker tomorrow. he's been bugging me with this for the past 2 days. and i keep telling him all i know is that x=2.

Short N Buff
02-01-2002, 05:35 PM
You have to memorize the powers of 2...3 or 4 to find log base whatever. such as 2^x=4. u just have to KNOW it. u cannot find it by hand. but we can find the sqrt of 5 w/out a calc.

(warning this is off topic, but i bring it up cuz its related)
you use a series of formulas, called the newtons method. it is used to determine the numerical value of a root function. its very easy but i don't remember it cuz no one is expected to memorize it. if you want me to look up the metod i'll do it and post it. (interesting note: the newtons method is used to determine radicals in a standard pocket calculator. the calculator doesn[t know sqrt(5) immediately. it has to go thru a series of repeating steps)

Podium Kreatin
02-01-2002, 05:36 PM
well, ln and log wouldn't really matter, b/c we're not dealing w/ neither powers of 10, or e. the properties would be teh same

also, for the 256 w/o showing work, generally, u can memorize a lot of 2^x's. if u use computers a lot, then u will notice them b/c they are binary numbers too (ie, videogame ssytems; 8-bit, 16-bit, 32-bit, 64-bit, 128-bit systems, etc):

f(x)=2^x
x- 0,1,2,3, 4, 5, 6, 7, 8
f(x)1,2,4,8, 16,32,64,128,256

Tryska
02-01-2002, 05:38 PM
well theoretically...my guy doesn't have to know beyond 256.

or the original 2^x-2=4

equation.

apparently....2's and the 4s represent specific bits, that will aid in creating the specific IP address he would need to get within the subnet. and ips only go up to 255.

Podium Kreatin
02-01-2002, 05:44 PM
wel, when i meant binary, i meant numbers

i have almost no memory of how to do this, cuz th last tiem i did this was in elementary school, but:
14= 0111 (look at th chart i made, 0x1 + 1x2 + 1x4 + 1x8, first digit is 2^0, second 2^1, etc)
or 07, 013, etc

Tryska
02-01-2002, 05:52 PM
yeah...i know binary quite well actually. my screen name is based on binary.

the way i learned was as follows:

think of the places in our decimal system, but plug em in by 2 and it's powers...

so
8421

and then add the numbers in from there...

ie:
0001 - 1
0010 - 2
0011 - 3
0100 - 4

the_hall
02-01-2002, 08:14 PM
x can't equal 8.

the problem is 2^x-2=2. If you were to say x is equal to 8, you would be saying that 2^8=4, or 256=4. 256 does not equal 4.

x is equal to 2, and only two. Here is how you would solve it, it is a simple algabraic problem, it can be solved using simple algebra, no "advanced" theorems.

2^x-2=2
2^x=4

Then you take the "x" root of both sides, so it would then look like this...

2= (the x'th root of 4)
When taking the root of something, you must realize that there is a positive answer, and a negative (especially since you know it is 2 [which is actually the square root])

so you have 2= +-(the x'th root of 4)

You know that for it to be equal to two, you have to take the square root of four, positive and negative.

So you end up with x=2,-2

does 2^-2 =4? No

does 2^2 =4? Yes

Mystic Eric
02-01-2002, 11:37 PM
Originally posted by the_hall
x can't equal 8.

the problem is 2^x-2=2. If you were to say x is equal to 8, you would be saying that 2^8=4, or 256=4. 256 does not equal 4.

x is equal to 2, and only two. Here is how you would solve it, it is a simple algabraic problem, it can be solved using simple algebra, no "advanced" theorems.

2^x-2=2
2^x=4

Then you take the "x" root of both sides, so it would then look like this...

2= (the x'th root of 4)
When taking the root of something, you must realize that there is a positive answer, and a negative (especially since you know it is 2 [which is actually the square root])

so you have 2= +-(the x'th root of 4)

You know that for it to be equal to two, you have to take the square root of four, positive and negative.

So you end up with x=2,-2

does 2^-2 =4? No

does 2^2 =4? Yes

buddy, if you took your time to actually read this thread, you would see that x=8 IS FOR A TOTALLY DIFFERENT EQUATION!!

the first equation is: 2^x - 2 = 2

the second and totally different equation is: 2^x - 2 = 254

so before you doubt my math skills again, read the equation

:rolleyes: yeeesh, some people's kids.

Tryska
02-02-2002, 05:06 AM
baha @some people's kids....

McBain
02-02-2002, 06:12 AM
tryska, im glad to see its all sorted, erics way is the way id do it, ie forget about all this log crap, try keep it as simple as possible, hope the dum guy does well.

the_hall
02-02-2002, 08:18 AM