You should be able to solve this

Desty · May 25, 2016, 10:34:27 AM

http://www.matteboken.se/media/571304/ma2b-vt15_red.pdf

Second problem a and b

Coomer · May 25, 2016, 10:36:21 AM

This is a bad thread

Desty · May 25, 2016, 10:38:50 AM

Don't feel too bad if you can't solve it. Just means you gotta dig your nose into a math book :^)

Elegiac · May 25, 2016, 10:41:57 AM

I got the fuck out of there so fast, I've never seen so many things I couldn't decipher all together at once

Desty · May 25, 2016, 10:42:06 AM

alright, I'll solve b and you do a

B) x=8
2x2=4
2x4=8

See? It's super simple

Desty · May 25, 2016, 10:42:54 AM

Literally first grade high school math for STEM

Ian · May 25, 2016, 10:45:27 AM

A. kind of stumps me, it's like the three fourths power or two thirds, somewhere in there. Am I supposed to use decimal for that? B. is 8.

Desty · May 25, 2016, 10:46:31 AM

Spoiler

guys, come on...

Ian · May 25, 2016, 10:49:02 AM

I haven't taken a high level math course in almost two years and I have zero interest in STEM when I eventually go back to college.

Quote from: Desty on May 25, 2016, 10:46:31 AM

guys, come on...

Desty · May 25, 2016, 10:51:08 AM

YouTube

1:04 100% me right now

Coomer · May 25, 2016, 10:58:44 AM

Quote from: Desty on May 25, 2016, 10:38:50 AM

Don't feel too bad if you can't solve it. Just means you gotta dig your nose into a math book :^)

Well seeing how I don't go on forums to do math, I'll decline thanks

Desty · May 25, 2016, 11:00:29 AM

Okay then... Whatever you say

Dietrich Six · May 25, 2016, 11:05:02 AM

Other languages are fun

Desty · May 25, 2016, 11:07:14 AM

Quote from: Dietrich Six on May 25, 2016, 11:05:02 AM

Other languages are fun

that's why I'm on this board, min vän

Also, Che khabar

Desty · May 25, 2016, 11:11:45 AM

x= lg5/lg3

>you couldn't solve it

maverick · May 25, 2016, 11:17:37 AM

Quote from: Desty on May 25, 2016, 11:11:45 AM

x= lg5/lg3

>you couldn't solve it

No work shown. Zero points.

Winy · May 25, 2016, 11:25:36 AM

a). 5^x = 3

You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.

Desty · May 25, 2016, 11:33:01 AM

Quote from: Winy on May 25, 2016, 11:25:36 AM

Spoiler
a). 5^x = 3
You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.

Good job, Winy! Here's the follow-up problem

10^-x
Clue: Log x = 0

Desty · May 25, 2016, 11:37:11 AM

Quote from: Winy on May 25, 2016, 11:25:36 AM

Spoiler
a). 5^x = 3

You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.

I will change my name to whatever you want for 5 days if you solve and show your work for the next problem I wrote down

🍁 Aria 🔮 · May 25, 2016, 11:37:59 AM

2 + 2 = 4

Boom, served

Winy · May 25, 2016, 11:39:20 AM

Quote from: Desty on May 25, 2016, 11:33:01 AM

Quote from: Winy on May 25, 2016, 11:25:36 AM
Spoiler
a). 5^x = 3
You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.
Good job, Winy! Here's the follow-up problem

10^-x
Clue: Log x = 0

I'm confused, mostly because I can't actually read the problem in the work sheet.

10^-x... Equals what? I'm not sure what it's asking.

Desty · May 25, 2016, 11:39:50 AM

Quote from: Prime Kruphix on May 25, 2016, 11:37:59 AM

2 + 2 = 4

Boom, served

I don't care; I'm still free
You can't take the sky from me

Desty · May 25, 2016, 11:42:23 AM

Quote from: Winy on May 25, 2016, 11:39:20 AM

Quote from: Desty on May 25, 2016, 11:33:01 AM
Quote from: Winy on May 25, 2016, 11:25:36 AM
Spoiler
a). 5^x = 3
You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.
Good job, Winy! Here's the follow-up problem

10^-x
Clue: Log x = 0
I'm confused, mostly because I can't actually read the problem in the work sheet.

10^-x... Equals what? I'm not sure what it's asking.

It says calculate 10^-x if log x = 0

Winy · May 25, 2016, 11:45:12 AM

Quote from: Desty on May 25, 2016, 11:42:23 AM

Quote from: Winy on May 25, 2016, 11:39:20 AM
Quote from: Desty on May 25, 2016, 11:33:01 AM
Quote from: Winy on May 25, 2016, 11:25:36 AM
Spoiler
a). 5^x = 3
You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.
Good job, Winy! Here's the follow-up problem

10^-x
Clue: Log x = 0
I'm confused, mostly because I can't actually read the problem in the work sheet.

10^-x... Equals what? I'm not sure what it's asking.
It says calculate 10^-x if log x = 0

Oh, duh, I should've been able to figure that out on my own lmao.

So Log(x) = 0

And that means x is just 1.

So 10^-x becomes 10^-1, which is just 1/10.

Desty · May 25, 2016, 11:55:48 AM

Spoiler

Super Irish · May 25, 2016, 01:21:44 PM

But do you know what an Erg is?

Ender · May 25, 2016, 01:28:40 PM

Ender · May 25, 2016, 01:37:20 PM

Quote from: Winy on May 25, 2016, 11:25:36 AM

a). 5^x = 3

You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.

beat me too it

Desty · May 25, 2016, 01:39:33 PM

Quote from: Ender on May 25, 2016, 01:37:20 PM

Quote from: Winy on May 25, 2016, 11:25:36 AM
Spoiler
a). 5^x = 3

You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.
beat me too it

Oh my god, what is going on

Is it international mispelling day or something? Things are starting to feel surreal

Winy · May 25, 2016, 01:42:53 PM

Quote from: Desty on May 25, 2016, 01:39:33 PM

Quote from: Ender on May 25, 2016, 01:37:20 PM
Quote from: Winy on May 25, 2016, 11:25:36 AM
Spoiler
a). 5^x = 3

You just take the log of both sides and solve for x

x * Log(5) = Log(3)

So x is Log(3)/Log(5)

If they want an approximate decimal, apparently that's about 0.6826

b). x^1/3 = 2

Just take the 1/3 root of 2, which is just the first root of 2³

So it's just 8.
beat me too it
Oh my god, what is going on

Is it international mispelling day or something? Things are starting to feel surreal

*Misspelling