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Mathematics: Post your doubts here!

scouserlfc

scouserlfc

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Can anyone solve this without using a calculator and giving the exact answer for x

(3x)^(lg3) = (4x)^(lg4)

I know its easy and even i solved it but can anyone tell me the answer in the exact form and if possible the steps u used to get them so i can compare to mine as i dont have the answer for this :p :p And a big thanks if u do solve it :D
 
P

PhyZac

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scouserlfc said:
Can anyone solve this without using a calculator and giving the exact answer for x

(3x)^(lg3) = (4x)^(lg4)

I know its easy and even i solved it but can anyone tell me the answer in the exact form and if possible the steps u used to get them so i can compare to mine as i dont have the answer for this :p :p And a big thanks if u do solve it :D
Click to expand...
it is very hard

lg [(3x)^(lg3)] = lg [(4x)^(lg4)]
lg3 x lg3x = lg4 x lg4x
lg3 x (lg3 + lgx) = lg4 x (lg4 + lgx)
(lg3)^2 + lg3lgx = (lg4)^2 + lg4lgx
(lg3)^2 - (lg4)^2 = lg4lgx - lg3lgx
(lg3 - lg4) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg4 -lg3) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg3 + lg4) = lgx
-lg12 = lgx
lg(1/12) = lgx
x = 1/12
 
iKhaled

iKhaled

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PhyZac said:
it is very hard

lg [(3x)^(lg3)] = lg [(4x)^(lg4)]
lg3 x lg3x = lg4 x lg4x
lg3 x (lg3 + lgx) = lg4 x (lg4 + lgx)
(lg3)^2 + lg3lgx = (lg4)^2 + lg4lgx
(lg3)^2 - (lg4)^2 = lg4lgx - lg3lgx
(lg3 - lg4) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg4 -lg3) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg3 + lg4) = lgx
-lg12 = lgx
lg(1/12) = lgx
x = 1/12
Click to expand...
is this statistics or math? lg is log? :S:S:S
 
Esme

Esme

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19islandprincess96 said:
Yes... but how do we get π/2?
Click to expand...

Okay..let's do it by another method then. We're given u, aren't we?
Now square the complex number, solve it and use that to find the argument.

I've forgotten the question, so if you can tell me what is u, I'll try and see if I can get the correct argument by this method.
 
SararaIH

SararaIH

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AlishaK said:
xdy/dx=1-y^2
rearrange so that all the x terms are at one side and y terms at the other.
so.... 1/(1-y^2) dy = 1/x dx
Now integrate both the sides,
$-sign for integration
$ 1/(1-y^2) dy = $1/x dx
P.S (1-y^2) is in the form a^2 - b^2 which is nothing but (a+b)(a-b)(remember??!)
so in order to integrate expressions like $1/a^2 - b^2 = [ 1/2a* ln(a+b/a-b)] (memorise this)
Therefore,
continuing with the integration,
$ 1/(1-y^2) dy = $1/x dx
[1/2*1 * ln(1+y/1-y)] = lnx +K where K is the common constant! :D
You're given: x=2, y=0
so put those in the eqn:
1/2 * ln (1+0/1-0) = ln2 +K
ln(1) is '0'.
therefore, K+ln2 = 0
K= -ln2 :DD
Now get the eqn
1/2 * ln(1+y/1-y) =lnx -ln2
1/2 * ln(1+y/1-y) -lnx + ln2 = 0

There u gooo!!
I hope u got it!! it can be easy...if u didnt understand then refer to the website i posted! u'll get it forever stuck in ur mind!!

Cheers!
Click to expand...
Hey, could you please tell me the final answer for the expression of 'y in terms of x'. The answer isn't written on the Mark Scheme :/
 
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19islandprincess96

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Esme said:
Okay..let's do it by another method then. We're given u, aren't we?
Now square the complex number, solve it and use that to find the argument.

I've forgotten the question, so if you can tell me what is u, I'll try and see if I can get the correct argument by this method.
Click to expand...

u= 2/(-1+i)
I'm sorry, I must be really annoying...
 
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