solve for $x$ : $4^x = x^4$

Asked Jan 01 '21 at 04:08

Active Jan 01 '21 at 07:47

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how to solve $x$ in $4^x = x^4$

My try: $$ 4^x = x^4$$ $$ x \ln4 = 4 \ln x $$ $$ 2x\ln2 = 4 \ln x $$ $$ \frac{1}{2} x \ln2 = \ln x $$ $$ e^{\frac{1}{2} x \ln2} = x ....(1)$$ $$$$ $$\text{Let} : u = \frac{1}{2} x \ln 2 ....(2)$$ $$\text{then}, x = \frac{2u}{\ln 2} ....(3)$$

sub (2) and (3)into (1): $$e^{u} = \frac{2u}{\ln 2}$$

And I have no idea of how to go on ...

edited Jan 01 '21 at 07:47

Albus Dumbledore

11,860

asked Jan 01 '21 at 04:08

user868733

you can immediately get $x=2$ as a solution just by inspection. Though, the question of whether this is the only solution is more involved (I think it is unique but I'm not too sure at the moment) – peek-a-boo Jan 01 '21 at 04:11
1

Google "lambert w-function" – morrowmh Jan 01 '21 at 04:14
$ x = 4 $ is also the answer – user868733 Jan 01 '21 at 04:14
Why is this tagged natural transformations? – Reveillark Jan 01 '21 at 04:14
I think the linked question is a good place to look - if you want to quickly connect it to your work, note that your equation rewrites as $$\frac{\ln(2)}2 = ue^{-u}$$ or, if $v=-u$, as $$\frac{-\ln(2)}2 = ve^v$$ which is the step at which you need the product log - but look at the linked question, it's more detailed about this. – Milo Brandt Jan 01 '21 at 04:17
@UtkarshJha ah yes, of course I missed the even more obvious solution $x=4$. – peek-a-boo Jan 01 '21 at 04:17
it helps, thx @Milo Brandt – user868733 Jan 01 '21 at 04:18

solve for $x$ : $4^x = x^4$

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