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

[bankebihari]

Originally Posted by pavanpadekal

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

yah..thats what..i think u cant take suare root because of inequality because it makes one side imaginary...cheer

[avenger007]

log(2ab)=loga*logb=log2+loga+logb-----1
log(bc)=logb*logc=logb+logc-------------2
log(2ac)=loga*logc=log2+loga+logc-----3

1-3
log(b/c)=log((b^loga)/(c^loga))
ie, b/c=(b/c)^loga
=>log a=1
hence base is a.
now,replace loga in (1) with 1,log(2ab)=logb=log2+1+logb
log 2=-1 for base a hence a=1/2;

now I'm stuck...wat nxt..help me..

[ravishah17]

Considering
Log a=p
Log b=q
Log c=r

Log2+p+q=p*q

q+r=q*r

Log2+p+r=p*r

q-r=p*(q-r)=> p=1; considering q-r not equal to 0 or q not equal to r
But this is not possible as we put value of p in eqn 1 & eqn 3

so q=r, 2q=q*q=> q=2 or q=0

Now for q=2; p=2+log2, r=2
for q=0, p=-2-log2, r=0
Putting back values of p,q,r
we have two solutions, answer is option (3)2

[ravishah17]

Originally Posted by avenger007

log(2ab)=loga*logb=log2+loga+logb-----1
log(bc)=logb*logc=logb+logc-------------2
log(2ac)=loga*logc=log2+loga+logc-----3

1-3
log(b/c)=log((b^loga)/(c^loga))
ie, b/c=(b/c)^loga
=>log a=1
hence base is a.
now,replace loga in (1) with 1,log(2ab)=logb=log2+1+logb
log 2=-1 for base a hence a=1/2;

now I'm stuck...wat nxt..help me..

I think you have cancelled some term in 1st step, assuming it to be not equal to zero, so answer is not possible... but that term would be equal to zero, try moving ahead from there... check out my solution....

[slam]

Hello people, here's my approach:
take loga = x, logb = y and logc = z
we get three equations in three variables:
2 + x + y = xy
y + z = yz
2 + x + z = xz

from the second we get z = (y)/(y-1)
from the first we get x = (y+2)/y-1)

put these in the third to get y(y-2) = 0
y=0 gives z=0 and x=-2 which gives us one set (a,b,c)
y=2 gives z=2 and x=4 which gives us another set (a,b,c)

Hence we get two sets of abc. => option(3).

I still have a bit of a question here. It has not been mentioned to what base we are taking the log of a,b and c.
Do we assume it to be 10 or e? My point here is that if we do not take it to be a fixed value, then from the values of (x,y,z) that have been calculated above, I could keep taking different bases to get different sets (a,b,c) which satisfy the three given equations.

Please comment/explain. Appreciate your help.

----------
slam.

[avenger007]

thanks..understood my mistake...

[avenger007]

Originally Posted by ravishah17

I think you have cancelled some term in 1st step, assuming it to be not equal to zero, so answer is not possible... but that term would be equal to zero, try moving ahead from there... check out my solution....

Thanks...understood my mistake

[vivek417]

Here is my solution:-

Let Loga =x, logb =y, logc =z;

so log2 + x+y = xy ----(1)
y+z=yz -----(2)
log2+x+z = xz -----(3)

(1) - (3) gives u, x=1 or y=z

But x=1 doesn't satisfy eqn 1

So y=z

Put y=z in eqn 2, we get y=0 or y=2

Using these values we get x = -log2 or log2 + 2

So a = 1/2 or a=200

Therefore the solution is (200,100,100) & (1/2,1,1)

So 2 solutions option 3.

[v-factor]

log(2ab) = log a * log b
=>log a + log b - log a * log b = -log2-------(1)

log(bc) = log b * log c
=>log b+ log c - log b* log c = 0-------------(2)

log(2ac) = log a * log c
=>log c + log a - log c * log a = -log2-------(3)

From equations (1) and (3)
log b - log c = log a + log b - log c * log a
=> Either Log a = 1 or b=c
Log a = 1 does not satisfy eq.(1)
Putting b = c in equation (2) we get log b = 0 or log b = 2
Thus we get 2 sets of values for (a,b,c)

Hence the option 3

[fiction]

@ aarav


I think my solution for yesterdays solution was also correct.