For what value of $'a'$, $\int_{0}^{2}a^xdx=3$? I solved further and got $$\frac{a^2-1}{\ln(a)}=3\Rightarrow a^2-1=3\ln(a)$$ $\forall a\ne1$. Putting this in wolfram alpha gives $a=1.464...$Is there any way to get this without using wolfram alpha?
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2You can't solve the equation $a^2-1=3\ln(a)$ with anything other than numerical methods. If you are asking for how to do that by hand then there are many such ways, maybe start with the bisection method? – QC_QAOA Aug 25 '23 at 16:51
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@QC_QAOA kk thnx. – Sam Aug 25 '23 at 16:53
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You can get a Lambert W solution and maybe a series/integral representation of it. – Тyma Gaidash Aug 25 '23 at 16:53
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@QC_QAOA I was able to find a Lambert W representation if you're curious – Ninad Munshi Aug 27 '23 at 06:01
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The equation can be rearranged to
$$a^2 = \frac{3}{2}\log a^2 + 1$$
or
$$-\frac{2}{3}e^{-\frac{2}{3}} = -\frac{2}{3}a^2e^{-\frac{2}{3}a^2}$$
The principal branch of Lambert W would give the trivial solution of $a=1$. However for negative numbers on the interval $(0,-e^{-1})$, there is another real solution with the $-1$ branch of the product log. This gives us the solution
$$W_{-1}\left(-\frac{2}{3}e^{-\frac{2}{3}}\right) = -\frac{2}{3}a^2 \implies \boxed{a = \sqrt{-\frac{3}{2}W_{-1}\left(-\frac{2}{3}e^{-\frac{2}{3}}\right)} \approx 1.46425_\cdots}$$
taking the positive square root since the base of the original integrand has to be a positive number.
Ninad Munshi
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