roshni

• Jul 8th, 2006

 

log 3.143/log3.142=log0.001;

log3.141=log3.142/log0.001

sumit.manchanda

• Mar 2nd, 2007

 

sorry to say but your answer is not correct
 becuse 

log (m/n)= logm- log n;

 but not  log (m-n) = logm/ logn
 thus how can u say that

 3.143- 3.142 = 0.001
taking log on both sides
log( 3.143-3.142) = log 0.001
 thus it is not like log(3.143-3.142)= log(3.143)/ log (3.142)

Srikanth B Pai

• Jul 9th, 2007

 

Hi all

Most likely one has to approximate.......
that is
let log3.142 = x, log(3.143) = y;
then generally speaking

log ((x-1)/1000) + log((x+1)/1000) = log(x^2 /1000^2) %%neglecting "+ 1/1000^2"

thus log 3.141 + log 3.143 = log((3.142)^2) %% x = 3142
So

ans = 2x - y

saket prakash

• Jul 18th, 2007

 

Here given
log 3.142=x, log 3.143=y;
10^x=3.142----------eqn 1
10^y=3.143----------eqn 2
Subtracting eqn 2-1 ;
.001= 10^y-10^x
and again subtracting
3.142-.001 =3.141
the answer is log 3.141=log( 10^x-(10^y-10^x))
log 3.141=log(2*10^x-10^y)

That's all
Thanking.you.

saksham

• Jul 27th, 2007

 

Yeah but in that case you need to know the value of log2... are we given that?

anuj_1497

 Profile Answers by anuj_1497 

• Jan 28th, 2008

 

subtract the value of log 3.142 from the value of log 3.143  and then subtract from the value of log 3.142 if the value from log 3.142 to log 3.143 is increasing  or subtract as if the value is decrease on going frm log 3.142 to log 3.143

hmvasista

 Profile Answers by hmvasista 

• Apr 29th, 2010

 

Log indentities are:
log a - log b = log(a/b)
log a + log b = log(ab)

we can proceed by taking let
log 3.142 = x, so 3.142= 10^x
log 3.143 = y, so 3.143= 10^y

and make appropriate substitutions....

prititripathi

 Profile Answers by prititripathi 

• Feb 25th, 2011

 

by applying log formula 

logm/logn=logm-logn

n log(3.143)/log(3.142)

=log.0001

again

log(3.142-.001)=log 3.141