CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[pavanpadekal]

Originally Posted by sabsebadapaagal

I have highlighted the portion where you have made mistake ....

it should a * loga

Thats is a typo..Next line is proper..Do tell me if i can use this approach to arrive at the answer

[implex]

Originally Posted by pavanpadekal

Put b=ka and c=pb
log(2ab)=log a*log b
log(2ka^2)=log a*log(ka)
log2+log k+2loga=log a*log k+log a^2
(log a)^2+log a(log k-2)-log k -log 2=0
log a =[-(log k -2)+-sqrt((log k -2)^2+4(log k+log2))]/2

Now for this to give values for log a from 0 to infinity
((log k -2)^2+4(log k+log2)>=0
((log k -2)^2>=-4(log k+log2)
(log k)^2+4-4log k>=-4log k-4log 2
(log k)^2>=-4(1+log 2)
(log k)^2>=-4(1.693)
log k>=sqrt(-6.772)



log(bc) = log b*log c
log(p*b^2)=log b * log p+ (log b)^2
log p+ 2 log b=log b * log p +(log b)^2
(log b)^2 +log b(log p-2)-log p=0
log b = [-(log p -2)+-sqrt[(log p-2)^2+4log p]/2
(log p-2)^2+4log p>=0
(log p) ^2+4>=0
(log p )^2>=-4
log p>=sqrt(-4)

subtracting the two equations we get
log(2a/c)=logb.log(a/c)...(1)
log(b/c)=loga.log(b/c)...(2)
log(2a/b)=logc.log(a/b) clearly the equations are symmetric in b and c so b=c


log(2a/c)=logb .Log(a/c)
log(a/c)+log 2=logc .log(a/c)
log(a/c)(logc-1)=log2
a/c=2 and log c=2
logc-1=log2 and loga/c=1
gives logc/2=1and loga/c=1

clealry we have two values

option 3)

[bankebihari]

Originally Posted by sabsebadapaagal

I have highlighted the portion where you have made mistake ....

it should a * loga

humm..log a*log(ka)=loga(logk+loga)=loga*logk+(loga)^2...so he was write at this stage...

he has done blunder in last stage..
(log k)^2>=-4(1.693)
log k>=sqrt(-6.772)....because square of any real number is always greater than any negative real number..so if this statement is write think where we r leading to....2nd statement is not saying same thing as 1st one..isnt it???

[bankebihari]

Originally Posted by pavanpadekal

Thats is a typo..Next line is proper..Do tell me if i can use this approach to arrive at the answer

pavan this is very lengthy and error prone approach...and ur concluding step is wrong..see my above post..

[pavanpadekal]

Originally Posted by bankebihari

humm..log a*log(ka)=loga(logk+loga)=loga*logk+(loga)^2...so he was write at this stage...

he has done blunder in last stage..
(log k)^2>=-4(1.693)
log k>=sqrt(-6.772)....because square of any real number is always greater than any negative real number..so if this statement is write think where we r leading to....2nd statement is not saying same thing as 1st one..isnt it???

I didnt follow buddy!

[bankebihari]

Originally Posted by pavanpadekal

I didnt follow buddy!

2^2>=(-3)..so according to ur step
2>=squrt(-2)..now this is meaningless since we cant compare a real number with an imaginary number...got me???

[rajeev_hts]

Sorry, I put my answer too early.

log(2a*b)=loga*logb --1
log(2a*c)=loga*logc --2
log(b*c)=logb*logc --3
From: 1/2
log(2ab)/log(2bc) = logb/logc
=> log(c/b)*log(2a) = 0
a=1/2 or b=c or both
but from 1
log(2*(1/2)*b)=log(1/2)*logb
This is wrong
so a=1/2 and b=c
or b=c only
for b=c
2logc= (logc)^2
logc=0,2
c=1 or 100
for c=1
b=1, a=1/2
for c=100
from 1
log200 + log a = log100 * loga
loga = log 200
a=200
b=100
so (a,b,c) are (1/2, 1, 1) and (200, 100, 100)

Answer (3)

[varunpriya]

why can't it be like this:

let log a = x; logb=y & logc=z
Given; log2+x+y=xy---1
y+z=yz----2
log2+x+z=xz----3

From 2; z=y/(y-1)
substituting in 3; log2+x+y/(y-1)=xy/(y-1)
which comes out to be(x-y)/(y-1)=log2---4
Substituting 4 in 1
x+y=xy-log2
x+y=xy-(x-y)/(y-1)
which comes out to be ; xy-2x=y-2
giving x=1 & y=2
therefore from 2, z=2
hence a=10,b=100 & c=100

only one solution hence 2 is the right choice

[pavanpadekal]

Originally Posted by bankebihari

2^2>=(-3)..so according to ur step
2>=squrt(-2)..now this is meaningless since we cant compare a real number with an imaginary number...got me???

But the problem is the sequence of my steps is exactly opposite.
I have proved
(log k)^2> =-(4(1+log 2))
and from that I'm deducing that
log k>=sqrt(-6.772)
i.e., equivalent to saying
x^2>=-4
and hence
x>=sqrt(-4)

Maybe I can't take square root because of the inequality sign.But the explanation you have given doesn't hold I believe

[krsh.vik]

let log a = a log b = b log c = c

we have three equations

a+b+2 = ab
b+c = bc
a+c+2 = ac

a= (b+2)/(b-1) c = b/(b-1)

substitute in a+c+2 = ac

we get b = 0 and b = 2

=> a = -2 c = 0 or a = 4 c = 2

log a = -2 log b = 0 log c = 0

or

log a = 4 log b = 2 log c = 2

two solutions