## [windows2000] Re: OT - Math Question

• From: Moby <moby@xxxxxxxxxxxxxx>
• To: windows2000@xxxxxxxxxxxxx
• Date: Wed, 29 Nov 2006 16:22:05 -0600

```

Chris Berry wrote:
```
```Moby wrote:

```
```Chris Berry wrote:

```
```Mark Lane wrote:

```
```I was a math major in a former life 10+ years.

1/2(e^x + e^-x) = 4
e^x + e^-x = 8
(e^x)^2 + 1 = 8e^x
Let y = e^x such that x = ln(y).
y^2 + 1 = 8y
y^2 - 8y + 1 = 0
```
Use quadratic formula, which is x = (-b +- sqrt(b^2 - 4ac))/2a, to solve for
```y in previous equation.  Then x = ln(y)
```
```

Pretty, but pure handwaving, you still didn't provide a value for x.

```
```To continue the handwaving some more then :)

```
Using the quadratic formula and keeping only the positive value for y (y = e^x - as such y cannot be negative!) we get:
```
y = (8 + 9.592) /2 = 8.796

This means e^x = 8.796 => x = ln(8.796) = 2.174
```
```
```
That doesn't work out right when I plug it into my spreadsheet. *scratches head*
```
I used:

=1/2*(2.71828183^A1+2.71828183^(A1*-1))

and got

4.4536 not 4

I'm a little rusty though, did I miss something?

```
It works out fine if you use more signigicant digits - I did all my calculations with only three significant digits, hence the non-exact match. Repeat my calculation with more significant digits and you will get a more accurate result.
```--

--Moby

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