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# factorising help!!! watch

1. how do i solve this:
ln2x - x = 0
if i factor x out, i get x(ln2 -1) 0
so i'm confused on how to solve the ln bit to get e^1/2??

EDIT : i worked out i made a mistake when factorising, it should have been x(lnx2 - 1) instead. For curiosity's sake, why isn't x is removed from lnx as well when factorising?
2. (Original post by elvss567)
how do i solve this:
ln2x - x = 0
if i factor x out, i get x(ln2 -1) 0
so i'm confused on how to solve the ln bit to get e^1/2??
Didn’t you ask the exact same question before and gave me rep for my answer? So it made sense...?
3. [QUOTE=RDKGames;74465352]Didn’t you ask the exact same question before and gave me rep for my answer? So it made sense...?[/QUOTE

i edited the question if you look^^
4. (Original post by elvss567)
how do i solve this:
ln2x - x = 0
if i factor x out, i get x(ln2 -1) 0
so i'm confused on how to solve the ln bit to get e^1/2??

EDIT : i worked out i made a mistake when factorising, it should have been x(lnx2 - 1) instead. For curiosity's sake, why isn't x is removed from lnx as well when factorising?

are not correct and you can not do this as ln(x) is an operator on x.

It is like doing sin(3x) implying x( sin(3) ), which IS NOT generally correct.

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As for your equation, it depends on interpretation, as we could have:

(note the modulus signs),

for which there is a solution but I do not think x can be found algebraically (anyone else can help here?).

I will give a tip that the exponential functions and ln are inverses of each other.
5. (Original post by simon0)

are not correct and you can not do this as ln(x) is an operator on x.

It is like doing sin(3x) implying x( sin(3) ), which IS NOT generally correct.

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As for your equation, note there are no real solutions (try graphing the function) but I will give a tip that the exponential functions and ln are inverses of each other.
got it! Thankyou

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Updated: November 7, 2017
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