Solving exponential equation with quadratic indicese2x2+3x−1−3ex2+1+5e2x+3=0

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Roy Brady · Accepted Answer

Step 1They are not indices but exponents.There is no way to find any analytical solution of this highly transcendental equation and numerical methods will be required.You could consider that you look for the zero&#039;s of functionf(x)=e2x2+3x−1−3ex2+1+5e2x+3but it so stiff that is is difficult so see where are, more or less, the solutions.Rearrange it, take logarithms and consider functiong(x)=log⁡(e2x2+3x−1+5e2x+3)−(x2+1)log⁡(3)which is reallmuch nicer. If you oberlap the plots ot the two logarithms, the problem &quot;looks&quot; like the intersetion of two parabolas showing solutions close to -4 and -1.Now, we are ready for a use of Newton method which, using these initial values, will generate the following iterates(nxn0−1.0000001−0.8792912−0.8756033−0.875599)(nxn0−4.0000001−3.8197222−3.8127173−3.812707)

Solving exponential equation with quadratic indices \(\displaystyle{e}^{{{2}{x}^{{{2}}}+{3}{x}-{1}}}-{3}{e}^{{{x}^{{{2}}}+{1}}}+{5}{e}^{{{2}{x}+{3}}}={0}\)

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