Natural numbers [please help]

[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?

[James R]

There is no solution.

[joeboo]

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

[Josh123]

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

[HallsofIvy]

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

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}}