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Math Problem

Name: Ein 2006-09-20 18:57

cant figure it out

^=exponent

2^2x - 33(2^x) + 32 = 0 solving for x

Name: Anonymous 2006-09-20 19:04

Come on, figure it out and I'll give you a gold star.

Name: Anonymous 2006-09-20 19:13

Replace 2^x with z, giving z^2-33z+32=0, which is easy to solve in your head: (z-32)(z-1)=0 -> z=32 or z=1.
So 2^x=32 or 2^x=1 -> x=5 or x=0.

Name: Anonymous 2006-09-20 19:18

*sticks a star to forehead*

Name: Anonymous 2006-09-20 20:03

Here's another, solve for x

2^2x - 33(2^x) + 32 = 0

Name: Anonymous 2006-09-20 20:05

^Oops, wrong one.

5^x^2 (625) = (1/25) ^2x

Name: Anonymous 2006-09-20 20:10

You need more brackets on the left, because I'm not sure what that's supposed to be. Also, on the right, ^ binds stronger than *, so you really should say (1/25)^(2x).

And do your own homework.

Name: Anonymous 2006-09-20 20:13

>>7
Thank you Mr. No-Help.

Name: Anonymous 2006-09-20 20:59

>>8
His points were valid: 5^x^2 (625) can be interpreted as (625)*5^(x^2) or (625)*(5^x)^2 or 5^(x^2 * 625) or 5^(x^(2*625)).

Name: Anonymous 2006-09-21 5:54

DIFFERENTIATE

Name: Anonymous 2006-09-22 20:22

>>9
Moreover, junior high pupils are not supposed to come to 4chan.

Name: Anonymous 2006-09-24 15:31

>>9
No, in the absence of brackets 5^x^2 (625) can only be interpreted as (625) * (5 ^ (x^2)). ^ always evaluates right to left.

Newer Posts

Math Problem

1 Name: Ein 2006-09-20 18:57

2 Name: Anonymous 2006-09-20 19:04

3 Name: Anonymous 2006-09-20 19:13

4 Name: Anonymous 2006-09-20 19:18

5 Name: Anonymous 2006-09-20 20:03

6 Name: Anonymous 2006-09-20 20:05

7 Name: Anonymous 2006-09-20 20:10

8 Name: Anonymous 2006-09-20 20:13

9 Name: Anonymous 2006-09-20 20:59

10 Name: Anonymous 2006-09-21 5:54

11 Name: Anonymous 2006-09-22 20:22

12 Name: Anonymous 2006-09-24 15:31