How can I solve 8n2=64nlog2(n)

Question

limacarp4 · Accepted Answer

64n⋅log2n−8n2=0⇒8n⋅(8log2n−n)=0 , hence : 8log2n−n=0 since n cannot be 0 , so: 8log2n=n⇒log2n=18⋅n⇒218n=n Substitute : 18n=t , so : 2t=8t⇒1=8t2t⇒1=8t⋅e−tln⁡2⇒18=t⋅e−tln⁡2⇒ ⇒−ln⁡28=(−t⋅ln⁡2)⋅e−tln⁡2⇒ ⇒W(−ln⁡28)=−tln⁡2⇒t=−W(−ln⁡28)ln⁡2⇒ ⇒n=−8⋅W(−ln⁡28)ln⁡2

Piosellisf · Accepted Answer

You could try to solve this using the Lambert W function but it is probably easier to do it numerically and find that (for integers)

f (n) < g (n)

when

2 \leq n \leq 43

.
For

n \geq 44

(and

n = 1

) you have

f (n) > g (n)

.

RizerMix · Accepted Answer

I assume you are only interested in positive values of n. The equation 8n2=64nlog2⁡(n) can be simplified to n=8log2⁡(n). WolframAlpha approximates the two solutions to this equation as n=1.1 and n=43.5593, and in between these two we have 8n2&amp;lt;64nlog2⁡(n) while otherwise 8n2&amp;gt;64nlog2⁡(n). Thus for 43≥n≥2 we have that g(n)&amp;gt;f(n), while otherwise (g(n)

How can I solve 8n^2=64n\log_2(n)

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