/sci/ - Science & Math

[Anonymous No.4959019]

Solve to EXACT VALUE (don't calculate that shit):

#1: [2^(x) - 2^(-x)] / 3 = 4
#2: [2^(x) - 2^(-x)] / 2 = 3
#3: [e^(x) - e^(-x)] / 2 = 4

I couldn't do it. Prove that I'm a retard for personal satisfaction.

[Anonymous No.4959060]

step by step:
1) use rules of logarithms
2) don't have down syndrome
3) ...
4) profit

[Anonymous No.4959068]

<div class="math">
\frac{2^x -2^{-x}}{3} &= 4
2^x -2^{-x} &= 12
</div>

It's magic.

[Anonymous No.4959080]

didn't bother to do them all, x=12/log6 for the first one

[Anonymous No.4959100]

openstudy.com

I discovered it last week. It's a great place to get help with your 7th grade homework that you are getting in college.

[Anonymous No.4959208]

>>4959019
Assuming x is real, then the solutions are
#1: x = ln(6 + √37)/ln 2
#2: x = ln(3 + √10)/ln 2
#3: x = ln(4 + √17)

[Anonymous No.4959263]

>>4959019
If you can solve one you can solve them all.
I'll show you how to do number 1:
2^x-2^(-x)-12=0
(2^x)^2-12*(2^x)-1=0 <--multiply by 2^x
u^2-12u-1=0 <---substitute 2^x=u
u=6-sqrt(37), 6+sqrt(37) <----quadratic formula
x=ln(6+sqrt(37))/ln(2) <---from 6+sqrt(37)=2^x, the 6-sqrt(37) doesn't have a real solution because it is negative.

And there you go, nothing a pre-calc student couldn't do. Now you can leave knowing you are retarded :)

[Anonymous No.4959286]

Observe that

<span class="math">\cosh x = \frac{e^x + e^{-x}}{2}[/spoiler]

<span class="math">arcosh x=\ln \left( x+\sqrt{x^{2}-1} \right)[/spoiler]

and also

<span class="math">2^x=e^{ln(2^x)}=e^{x*ln(2)}[/spoiler]

http://en.wikipedia.org/wiki/Hyperbolic_function#Standard_algebraic_expressions

[Anonymous No.4959420]

[Anonymous No.4959425]

[Anonymous No.4959428]