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I came along this equation in a book for higher education engineering:

$$ x\cdot2^x - x^2 = 135 $$ What I know is that there are ways to get a solution numerically. But I‘m pretty sure that there is also an analytical way (besides rather complicated stuff like lamberts function) since the result is (simply) $$ x=5$$ Did I miss a simple trick?

Best wishes!

Thorben
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    The trick is probably just to guess and check. – Vishu Mar 14 '21 at 11:51
  • Oh man! I mean, this would be disappointing – Thorben Mar 14 '21 at 11:59
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    Let $f(x)$= LHS - RHS : it is an increasing function with a unique explicited root $x=5$. Why would you need an analytic expression ? If we replace now $135$ by $134$, besides a possible expression with Lambert $W$ function, I am almost sure there is no analytic expression of the root. – Jean Marie Mar 14 '21 at 12:10
  • Well, $$2^5-5^2=7$$ $$2^7+2^5-5^2=2^7+7=135$$ $$(2^2+1)2^5-5^2=135$$ $$5\cdot 2^5-5^2=135$$ I guess that doesn't really help, although it might be obvious to someone who thinks in binary. Note that $x\cdot 2^x-135=x^2$, so $x\cdot 2^x\ge135$, which pretty quickly leads to $x>4$ – PM 2Ring Mar 14 '21 at 15:17

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I don't know if this is an upgrade over guessing but you can try this,

$x(2^x -x) = 135$

The factors of $135$ are $1 \times 135 , 3 \times 45,5 \times 27,9 \times 15$ so now you have a guess at x, So check for $(2^x -x)$ and at $x = 5 ,$ we get $(2^x -x) = 27$

Vinanth S Bharadwaj
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