haha

**dolor** · Dec 13th, 2003, 04:26 PM

If 2x+1 = 8, find the value of 4x+1.

**TheManWhoCan** · Dec 13th, 2003, 04:53 PM

Isn't that just 15?

**Acidic** · Dec 13th, 2003, 06:44 PM

that's what seems obvious, have we missed something? maybe you are using a different number system? Why would this thread be called haha?

**wossname** · Dec 14th, 2003, 09:22 AM

Its evidently a conspiracy that stems back to Roswell and JFK.

**NotLKH** · Dec 14th, 2003, 03:19 PM

Hmmm, if he meant to use [sup] tags, but somehow forgot, then it becomes more interesting:

If 2^x+1 = 8, find the value of 4^x+1.

50.

Thats the answer, now how's its done without a calculator?

**Acidic** · Dec 14th, 2003, 04:01 PM

Using a calculator I get the same answer, but using pen and paper I get 65.
2^x+1=8
can be written as:
2^x+2⁰=2³ right?
As they are all to the same base we can re-write it to:
x+0=3
x=3

putting that in the other eqn:
4³+1 = 65

confused :sigh:
Someone tell me where I went wrong.

**NoteMe** · Dec 14th, 2003, 06:23 PM

Originally posted by NotLKH

Thats the answer, now how's its done without a calculator?

Draw it on a pwace of paper....

....y=4^x +1

**NoteMe** · Dec 14th, 2003, 06:37 PM

Originally posted by Acidic
Using a calculator I get the same answer, but using pen and paper I get 65.
2^x+1=8
can be written as:
2^x+2⁰=2³ right?
As they are all to the same base we can re-write it to:
x+0=3
x=3

putting that in the other eqn:
4³+1 = 65

confused :sigh:
Someone tell me where I went wrong.

Your problem is that you are ignoring the +1. You think that you can write it as 2^0 that seems logical. But you can't. You first have to take it over to the other side. Like this.

2^x + 1 = 8

becomes

2^x = 7

then you can do as you did.

2^x = 2^2.807354922

then:

x = 2.807354922

then:

4^x + 1 = 50

UNderstand?

**jemidiah** · Dec 14th, 2003, 07:06 PM

Using logarithms:

2^x + 1 = 8
2^x = 7

Basic logarithm formula:
b^x = c
Ex. log(100) = 2 [Note: 'log' is called Common Logarithm which is in base 10.]
So, b = 10, c = 100, x = 2

So
2^x = 7 would make

log(7) in base 2 = log(7)/log(2)

So x = 2.8073549220576041074419693172318 like has been stated.

4^{2.8073549220576041074419693172318} + 1 = 50, again, like has been stated.

**NoteMe** · Dec 14th, 2003, 07:09 PM

Originally posted by jemidiah
Using logarithms:

2^x + 1 = 8
2^x = 7

Basic logarithm formula:
b^x = c
Ex. log(100) = 2 [Note: 'log' is called Common Logarithm which is in base 10.]
So, b = 10, c = 100, x = 2

So
2^x = 7 would make

log(7) in base 2 = log(7)/log(2)

So x = 2.8073549220576041074419693172318 like has been stated.

4^{2.8073549220576041074419693172318} + 1 = 50, again, like has been stated.

Yeah that is the calculator way to do it...

....but can you calculate Log in your head????

**jemidiah** · Dec 14th, 2003, 08:06 PM

I'd thought you meant you used a Solve function or similar on a calculator

As for logarithms without a calculator, I'd assume it's related to the roots question in that it most likely uses some converging series.

**NoteMe** · Dec 15th, 2003, 03:41 AM

Yeah probably. I have never seen Acidics way of doing it. But I liked it. With the right numbers it can be pretty easy to solve the problem.....

**sql_lall** · Dec 15th, 2003, 03:41 AM

2^x+1=8
=> 2^x = 7
=> 2^x * 2^x = 7 * 7
=> 4^x = 49
=> 4^x + 1 = 50 !!!

no calculator needed

BTW:
a^x+a^y=a^z doesn't mean x+y=z.
However, a^x*a^y=a^z does

**NoteMe** · Dec 15th, 2003, 03:44 AM

Wow...it looks like it works too...I am impressed sql_lall....

**Something Else** · Dec 15th, 2003, 05:55 AM

Originally posted by sql_lall
2^x+1=8
=> 2^x = 7
=> 2^x * 2^x = 7 * 7
=> 4^x = 49
=> 4^x + 1 = 50 !!!

Very Nice!

-Lou

**Acidic** · Dec 15th, 2003, 08:29 AM

very nice, but does it always work?

**NoteMe** · Dec 15th, 2003, 08:43 AM

Originally posted by Acidic
very nice, but does it always work?

Yeah it will always work. Because you have

2^x + 1 = 8

2^x = 7 <-- remember this

And you can always multiply with the same number on both side. So if what I wrote over is true the "number" 2^x = the number 7

So the multiply "7" on both side will give you.

2^x * 2^x = 7 * 7

That gives you:

2^2x = 49

You can rewrite that to

4^x = 49

And that you can rewrite to

4^x + 1 = 50

QED....

**Acidic** · Dec 15th, 2003, 08:46 AM

2^x = 7 <-- remember this

And you can always multiply with the same number on both side. So if what I wrote over is true the "number" 2^x = the number 7

So the multiply "7" on both side will give you.

2^x * 2^x = 7 * 7

Basically you squared both sides.

That is a very nice method.

**NoteMe** · Dec 15th, 2003, 08:50 AM

Originally posted by Acidic
Basically you squared both sides.

As you will fancy-smancy english writing UK guy....

J/K

**Acidic** · Dec 15th, 2003, 09:06 AM

Actually I'm Swedish, but now live and study in wales.
Though I will be going back to sweden for Xmas.

**NoteMe** · Dec 15th, 2003, 09:10 AM

Hehe...ok then...

....but I was thinking about it when I went to the toilet. And what I said is not the same thing as what you said. But the answer is the same. You are always allowed to do the same thing on both sides of the equation. And I multiplied "7" on both sides, and you squared it....the same answer, but not the same way....

**Acidic** · Dec 15th, 2003, 09:14 AM

anyway, all this is based on that Dolor forgot the [sup] tags.

**NoteMe** · Dec 15th, 2003, 09:18 AM

Originally posted by Acidic
anyway, all this is based on that Dolor forgot the [sup] tags.

Yeah....or did he???...

**kedaman** · Dec 15th, 2003, 03:47 PM

I don't want to be a hypocrit but you have too much time on your hand

**Acidic** · Dec 15th, 2003, 03:50 PM

Is that a bad thing?

**NoteMe** · Dec 15th, 2003, 03:55 PM

Originally posted by kedaman
I don't want to be a hypocrit but you have too much time on your hand

Who where you talking too.....I don't have that ....I have Database exam tomorrow....

**kedaman** · Dec 15th, 2003, 04:06 PM

note: and what have you been doing all day

acidic: dunno, but I had to put a disclaimer so i wouldn't be called hypocrit just in case

**NoteMe** · Dec 15th, 2003, 04:11 PM

Originally posted by kedaman
note: and what have you been doing all day

I have been:

SELECT K.hypocritIndex
FROM KEDA K
WHERE K.name = 'Jerk'
GROUP BY K.hypocritIndex
HAVE COUNT(*) > 99999999

ANd guess what came out.....

You did....

J/K

**kedaman** · Dec 15th, 2003, 04:18 PM

yeah try that in your exam note, and see what you get

SELECT note.exam WHERE note.score=0 FROM database.exam

**NoteMe** · Dec 15th, 2003, 04:21 PM

Originally posted by kedaman
yeah try that in your exam note, and see what you get

SELECT note.exam WHERE note.score=0 FROM database.exam

Hehe...you are funny....

Answer := project name, ***, country (select name = 'Keda' (VBForum product CCF));

output:

Keda, True, Finland.

**kedaman** · Dec 15th, 2003, 04:25 PM

whats those three stars for? btw I haven't taken the database course yet, so I haven't got a score yet

**NoteMe** · Dec 15th, 2003, 04:27 PM

Originally posted by kedaman
whats those three stars for? btw I haven't taken the database course yet, so I haven't got a score yet

*** = BØG = ****** = Gay = Homo...and so on....

**kedaman** · Dec 15th, 2003, 04:45 PM

Originally posted by NoteMe
*** = BØG = ****** = Gay = Homo...and so on....

yeah and you forgot the last thing.. = Note

**NoteMe** · Dec 15th, 2003, 04:57 PM

Yeah "Note that Keda is a bøg"....

**Acidic** · Dec 15th, 2003, 05:02 PM

or is he a bög maybe?
j/k

**NoteMe** · Dec 15th, 2003, 05:04 PM

Wow...Norway Sweden and Finland in the same thread....

...calling each other Bøg, Bög, Bøgis....

...this is great...

**Acidic** · Dec 15th, 2003, 05:09 PM

no terve. Minä haluan [****, I don't know icecream in finnish]
ok:
Hallo, Ich will Glass ******. (?)

**NoteMe** · Dec 15th, 2003, 05:12 PM

Je ne pas parle 'Germen"....

**Acidic** · Dec 15th, 2003, 05:16 PM

et je parle francais mais avec tres difficulte.

**NoteMe** · Dec 15th, 2003, 05:20 PM

Tu parle francais trés bien. Mais je ne meme pas ma francais.

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