Prove this? (1 Viewer)

[ayehann]

i know there probably some silly rule that says this but...
how do i prove that ln2 = -ln(1/2)

 

[MC Squidge]

LHS=ln2

RHS=-ln(1/2)
=-ln(2^-1)
=--1ln2
=ln2
=LHS

LHS=RHS
:. ln2=-ln(1/2)

 

[Timothy.Siu]

i know there probably some silly rule that says this but...
how do i prove that ln2 = -ln(1/2)

yeah, coefficeints can be brought up

RHS=-ln(1/2)=ln[(1/2)^-1]=ln2

 

[hermand]

since



since

 

[ayehann]

AHHH! thanks guys...
btw isnt it funny how your all doing extn 2 maths.. just noticed that

 

[kwabon]

since



since

pretty cool way

 

[micuzzo]

how about:

ln2=-ln0.5

ln2 + ln0.5 = 0

LHS = ln1=0=RHS

Q.E.D

 

[Timothy.Siu]

how about:

ln2=-ln0.5

ln2 + ln0.5 = 0

LHS = ln1=0=RHS

Q.E.D

i dont think thats a good proof lol.
u changed the question btw. but i guess u cud....
its best to work from one side to the other.

 

[hermand]

since



since

interesting to note that

which can be proven by the above method.

unless we're supposed to know that?

 

[ayehann]

since



since

interesting to note that

which can be proven by the above method.

unless we're supposed to know that?

not sure... i hadnt come across this until todayy when i was revising physical apps of calculus.. and i got an answer as -ln0.5 and the actual answer was ln 2

 

[jack_smith]

u should know that straight away imo

asif use working

 

[GUSSSSSSSSSSSSS]

how about:

ln2=-ln0.5

ln2 + ln0.5 = 0

LHS = ln1=0=RHS

Q.E.D

no im fairly sure you CANNOT do that
as that would involve ASSUMING that the equality holds

LHS and RHS should be kept separately at all times

 

[jet]

In all actuality, it is just an application of a log law which you can assume:

 

[hermand]

In all actuality, it is just an application of a log law which you can assume:

i thought we might have been supposed to know that in some form. eh. i do now. haha.

thanks.

 

[azureus88]

no im fairly sure you CANNOT do that
as that would involve ASSUMING that the equality holds

LHS and RHS should be kept separately at all times

nah, im pretty sure you can do that, he's just adding ln0.5 to both sites. If the original equality holds, then adding ln0.5 to both sides should preserve the equality.

 

[MC Squidge]

u should know that straight away imo

asif use working

k fuck off if you arent going to be helpful

 

[undalay]

Someone prove: loga + logb = logab

 

[GUSSSSSSSSSSSSS]

nah, im pretty sure you can do that, he's just adding ln0.5 to both sites. If the original equality holds, then adding ln0.5 to both sides should preserve the equality.

thats the problem....that argument relies IF the original equality holds

we have to PROVE that it holds, and therefore CANNOT assume that it holds in the first place

 

[MC Squidge]

u can fuck off to you smartass

edit: @ undalay

 

[Trebla]

Someone prove: loga + logb = logab

Being general with base x, consider:
y = x^{log_xa + log_xb}
= (x^log_xa)(x^log_xb)
= a.b
Therefore:
log_xy = log_x(ab)
Also,
log_xy = log_x(x^{log_xa + log_xb})
= log_xa + log_xb
So the formula holds...