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factorising help!!!

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?
(edited 6 years ago)
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...?
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Avatar for elvss567
elvss567
OP
10
[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^^
Reply 3
Avatar for simon0
simon0
13
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?


Your factorisations of:

ln⁡(2x)−x, \ln(2x) - x,

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.

----------------------------------------------------------------------------------------------------

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

ln⁡∣2x∣−x=0, \ln \lvert 2x \rvert - x = 0, (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.
(edited 6 years ago)
Reply 4
Avatar for elvss567
elvss567
OP
10
Original post by simon0
Your factorisations of:

ln⁡(2x)−x, \ln(2x) - x,

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.

----------------------------------------------------------------------------------------------------

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:smile:

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