If $x^\sqrt x=\frac{3x}{2}$ then how to solve it? I have tried by taking $\log$ in both side but couldn't reach a solution. Do you have any logarithmic or any other solution?
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By inspection, we can see that $\dfrac{9}{4}$ is one solution. You might be stuck with numerical methods to find the other. – Moo Jun 19 '19 at 20:19
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Try this substitution:
Let $y = \sqrt{x}$.
Then $y^{2} = x$, and so, the original equation transforms to
$(y^{2})^y = \frac{3y^{2}}{2}$,
which, after some algebraic manipulation, yields
$y^{2y - 2} = \frac{3}{2}$;
Upon inspection, we see that $y = \frac{3}{2}$;
Whence---
$x = \frac{9}{4}$.
DDS
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good step that of substituting (+1) : now at the second line we could try and take the log, maybe we get a step ahead .. – G Cab Jun 19 '19 at 22:34
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@G Cab Thanks for the suggestion. I tried that; I recall that it allowed me to explicitly derive $y =\frac{3}{2}$ and so $x=\frac{9}{4}$; but not, unfortunately, the more interesting solution. Thanks again. – DDS Jun 19 '19 at 23:14
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substitute in $y=3/2$ for both values of $y$ on the LHS and you get $3/2$ which equals the RHS – DDS Jun 20 '19 at 04:29
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Using Mathematica, by inspection there are two solutions. (There are no solutions for $x \geq 3$, because $x^\sqrt{x}$ grows much faster than $\frac{3}{2}x$. This is something you can prove, even though it might not be possible to solve for the roots.)
They look to be about 0.4 and 2.2, so using FindRoot with these guesses gives $\frac{9}{4}$ (as the above commenter said) and $0.361634$. By WolframAlpha, the latter solution does not have an obvious closed form.
aras
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Some of the correct ways to solve such an equation is to use a numerical iterative method such as Bisection Method or Newton's Method. The bisection method may be easier to use since the derivative required by the later method is rather unfriendly.
NoChance
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