[Thisisit]

Originally posted by Luke.

how about this one
log(x^2) - log(x^2-x) = 1

I'm assuming you mean natural logs? I usually write them as ln.

log(x^2) - log(x^2-x) = 1
= log[(x^2)/(x^2-x)] = 1

raise e to the power of each side

(x^2)/(x^2-x) = e
1/[1-(1/x)] = e
1-(1/x) = 1/e
1- 1/e = 1/x
x = (1-1/e)^-1

[TheRoofInhabitant]

Originally posted by Thisisit

I'm assuming you mean natural logs? I usually write them as ln.

log(x^2) - log(x^2-x) = 1
= log[(x^2)/(x^2-x)] = 1

raise e to the power of each side

(x^2)/(x^2-x) = e
1/[1-(1/x)] = e
1-(1/x) = 1/e
1- 1/e = 1/x
x = (1-1/e)^-1

logs are decimal logarithms, therefore 10 instead of e.

[Sombra]

Originally posted by TheRoofInhabitant

logs are decimal logarithms, therefore 10 instead of e.

Yeah

[Donquizote]

Originally posted by TheRoofInhabitant

logs are decimal logarithms, therefore 10 instead of e.

cORRECT. 10^1

[jax.]

Yesss!!!! I'll do this when I get home, I love math!

[rl231]

Wait, so does x = 10/9 with the log problem?

[TheRoofInhabitant]

Originally posted by Donquizote

cORRECT. 10^1

Originally posted by rl231

Wait, so does x = 10/9 with the log problem?

Yep

[Donquizote]

Edited:

So
(x^2)/(x^2-x)=10^1
x^2/ (x(x-1)=10

x/ (x-1)=10

x= 10x-10
-9x=-10

so x=10/9

[Thisisit]

Originally posted by XLR8

-1 + sin^2x + sec^2(-x)
__________________
cos


Shake ya ass to that one


Edit: LMAO@54 EVER being the lesser value to -4

ermm tanxsinx -tan^2(x)secx?

It's far too late in the evening to try and simplify this

[Sombra]

Originally posted by rl231

Wait, so does x = 10/9 with the log problem?

Yes!
9x^2 - 10x = 0
x = 0, 10/9
x = 0 is discarded due to not agreeing with the equation, therefor 10/9 is the solution

[Thisisit]

Originally posted by TheRoofInhabitant

logs are decimal logarithms, therefore 10 instead of e.

then just replace all the e's with 10, so (1-1/10)^-1 = (9/10)^-1 = 10/9

[Thisisit]

I just prefer working with e cause it's neater

[TheRoofInhabitant]

Originally posted by Thisisit

I just prefer working with e cause it's neater

I know the feeling, but lately I have been working with decimal logs in chem, and I pay attention to the notation, getting confused between the two has become pretty easy

[Thisisit]

Originally posted by TheRoofInhabitant

I know the feeling, but lately I have been working with decimal logs in chem, and I pay attention to the notation, getting confused between the two has become pretty easy

OMG yes I almost lost a mark on my chemistry test the other day cause I used log base e in a pH calculation spotted it at the last minute though

anyways, let's get some integration up in this bitch

∫ 1/(x^2 - 4x +7) . dx

[Sombra]

Originally posted by Thisisit

OMG yes I almost lost a mark on my chemistry test the other day cause I used log base e in a pH calculation spotted it at the last minute though

anyways, let's get some integration up in this bitch

∫ 1/(x^2 - 4x +7) . dx

=(-1/2)-2x^2+7x+C

[Celebrate]

ln(x^2-4x+7)*(1/(2x-4))+C

[Thisisit]

Originally posted by Luke.

=(-1/2)-2x^2+7x+C

Originally posted by Zune

ln(x^2-4x+7)*(1/(2x-4))+C

Nope and nope. Differentiating the top one will leave you with 4x+7, and the bottom will leave you with something ugly which I cba to work out/

[TheRoofInhabitant]

My connection has left me, I'm on my phone, anyway x^2-4x+7 can be seen as (x-2)^2+root(3)^2, so in the end the result is 1/root(3) arctan(x-2)/root(3) + c

[Thisisit]

BTW the correct answer is (1/sqrt3)arctan[x-2/(sqrt3)] + C

[Thisisit]

Originally posted by TheRoofInhabitant

My connection has left me, I'm on my phone, anyway x^2-4x+7 can be seen as (x-2)^2+root(3)^2, so in the end the result is 1/root(3) arctan(x-2)/root(3) + c

Slay.