1. Apr 21, 2005

### Josh123

Hello. I am working on this problem

0=0.002*e^-(0.005/2R)

I am supposed to find to find "R". The only way I know how to do this gives me 0... but I know that it's not the answer. Got any tips?

2. Apr 21, 2005

### James R

There is no solution.

3. Apr 21, 2005

### joeboo

e^(x) > 0 for every x. This should clarify the point made by James R

4. Apr 21, 2005

### Josh123

What if the number is small (but not zero)... ie 0.000001 = 0005e^(0.004/2R)

5. Apr 22, 2005

### HallsofIvy

Staff Emeritus
Then it's easy (and NOT "Calculus and Analysis"!). Divide both sides by 0.0005 to get $$e^{\frac{0.004}{2R}}= \frac{0.000001}{0.0005}$$.

Take the natural logarithm of both sides to get rid of the exponential:
$$\frac{0.004}{2R}= ln(\frac{0.000001}{0.0005})$$

Multiply both sides by R:
$$0.002= R ln(\frac{0.000001}{0.0005})$$
and, finally, divide both sides by the logarithm:

$R= \frac{0.002}{ln(\frac{0.000001}{0.0005}}$

Last edited: Apr 22, 2005