Some Fun With Maths

[CraigD]

d=7/(3e - 9*(3x-2)^4)

:thumbs_up That agrees with what I got - but you beat me to posting it, sanctus. In pretty LaTeX:

 

[math]d = \frac{7}{3(e -3(3x -2)^4)}[/math]

 

I was curious to see how long it actually takes me to do these manipulations (with my usual interruptions it seems like even simple things start and end hours or days apart), so I timestampted when I started and stopped, getting an elapsed time of 4 minutes, much slower than I would have guessed. I think I'm slower, because, as with nearly everything for the past couple of decades, I'm using a text editor rather than paper and pencil. Despite popular hype to the contrary, electronics don't always make things faster.

[Aethelwulf]

Right ok... let's check it over...

 

We differentiate

 

[math]\frac{d}{dx}[\frac{1}{5}(3x - 2)^5][/math]

 

We need to use the chain rule and the power rule

 

[math]5 \cdot \frac{1}{5}(3x - 2)^4 \cdot 3[/math]

 

[math]=3(3x - 2)^4[/math]

 

Now to the second part,

 

[math]\frac{3d}{d - 2}/ \frac{9d^2}{7d - 14}[/math]

 

We multiply by the reciprocal which involves a factor

 

[math]\frac{3d}{d - 2} \cdot \frac{7d - 14}{9d^2}[/math]

 

[math]\frac{3d}{d - 2} \cdot \frac{7(d - 2)}{9 \cdot d \cdot d}[/math]

 

so we have

 

[math]\frac{3 \cdot d \cdot 7 (d - 2)}{(d - 2) \cdot 9 \cdot d \cdot d}[/math]

 

The [math](d - 2)[/math]'s cancel out including one factor of [math]d[/math] leaving

 

[math]\frac{21}{9d}[/math]

 

We rewrite the equation then as follows

 

[math]a + b = e[/math] gives

 

Thus

 

[math]\frac{3(3x - 2)^4 + 21}{9d} = e[/math]

 

multiply [math]9d[/math] on both sides

 

[math]3(3x - 2)^4 + 21 = 9de[/math]

 

then divide [math]9e[/math] on both sides

 

[math]\frac{3(3x - 2)^4 + 21}{9e} = d[/math]

Edited December 14, 2012 by Aethelwulf

[Aethelwulf]

examples? sure.

 

4x²+12x+12=

solve for x={1,2,3...}

 

let x= 1:

4*1²+12*1+12=?

4*1+12+12=?

4+12+12=28

 

let x=2:

4*2²+12*2+12=?

4*4+24+12=?

16+24+12=52

 

let x=3:

4*3²+12*3+12=?

4*9+36+12=?

36+36+12=84

 

 

 

Ok looks good to me. :)

[Aethelwulf]

The value of [imath]a[/imath] isn't important.

 

 

 

Of course the value of a is required - you are giving a fraction, whether simple or not, and you are given a value to b as well. If you simplified the fractions like I had explained in the OP, requiring no less a little bit of algebra, one will find a much simpler, easier to deal with equation.

[sanctus]

[...]

[math]\frac{21}{9d}[/math]

Till here we agree, although I simplified your fraction to [math]\frac{21}{9d}=\frac{7}{3d}[/math]

 

We rewrite the equation then as follows

 

[math]a + b = e[/math] gives

 

Thus

 

[math]\frac{3(3x - 2)^4 + 21}{9d} = e[/math]

Here is the disagreement, you put the denominator of b also below a.

It should be:

[math]3(3x - 2)^4 + \frac{21}{9d} = e[/math]

 

which then leads to craigs and my result. Do you agree?

[Aethelwulf]

Emmm yes... I have made an error there, in which case, we should arrive at those conclusions. Well done :)

 

I will type up another soon.