Can You Solve the Nonlinear Equation v_0 = 100e^{-v_0/100}?

[transgalactic]

<br /> v_0=100e^{\frac{-v_0}{100}}\\<br /> \ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\<br />
the answer is v_0=56.7
how to find v_0
??

 

[Mark44]

You can't solve this type of equation by algebraic means, but you can use approximation techniques to get an approximate solution, which v₀ = 56.7 appears to be.

 

[Shooting Star]

<br /> v_0=100e^{\frac{-v_0}{100}}\\<br /> \ln {v_0}=\ln{e^{\frac{-v_0}{100}}}^{100}\\<br />
the answer is v_0=56.7
how to find v_0
??

This is one way of doing it.

<br /> v_0=100e^{\frac{-v_0}{100}}.<br />


Putting <br /> {\textstyle{{{\rm v}_{\rm 0} } \over {100}}} = x, x = e^{ - x} \Rightarrow x = 1 - x + x^2 /2,<br /> retaining up to the second order term after expanding e^{-x}.

Now solve for x to get the approximate value. Justify why you are neglecting one value.