I can't figure it out. =/ $\displaystyle ln(x) = x - 2$ $\displaystyle x = e^{x - 2}$ $\displaystyle x = \frac{e^x}{e^2}$ I'm quite stuck. I feel like I should know how to do this, but just can't remember. >_<
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You need to solve it numerically I'm afraid. Plotting it gives rough ideas of x= 0.16 and 3.13
How irksome. Thanks though.... >_< I detest having to use Newton's method.
Originally Posted by Zizoo How irksome. Thanks though.... >_< I detest having to use Newton's method. Use the Bisection Method then :P
Originally Posted by e^(i*pi) You need to solve it numerically I'm afraid. Plotting it gives rough ideas of x= 0.16 and 3.13 Can be solved in closed form in terms of Lambert's W function (real (0) branch for a real solution): $\displaystyle x=-\text{W}(-e^2)=0.158594$ CB
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