Elitmus
Exam
Log(xy^3)=a, Log(x^2y)=b, Find log(y)/log(x)
Read Solution (Total 9)
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- log(xy)=log x+log y
So log(xy^3)=log x+log y^3=log x+log y+log y+log y=log x+3log y=a--->(1)
log(x^2y)=log x^2+log y=log x+log x+log y=2log x+log y=b--->(2)
On solving equations 1 & 2 we get logx=(3b-a)/5,logy=(2a-b)/5
So log(y)/log(x)=(2a-b)/(3b-a) - 11 years agoHelpfull: Yes(41) No(3)
- a=log xy^3 => a= log x+ 3log y => log x= a- 3log y......(1)
b=log x^2y => b= 2log x+ log y => b= 2(a-3log y)+log y => b=2a- 5log y
log y=(2a-b)/5;
after putting value of log y in eq.(1) we get
log x = a-3(2a-b)/5=(3b-a)/5;
So Ans. log y/ log x =(2a-b)/(3b-a) - 11 years agoHelpfull: Yes(5) No(0)
- a=log xy^3 => a= log x +3log y;
b=log x^2y => b=2log x +log y;
a-b=2log y -log x => a-b=2 log y/x;
log y/x=(a-b)/2 - 11 years agoHelpfull: Yes(2) No(6)
- log(xy^3)=logx+3logy=a----1
log(x^2y)=2ylogx=b--------2
now,
multiply eqn 1 by 2y on both sides
2ylogx+3.2ylogx=2ay
sub eqn 1 by 2
we get logy=2ay-b/6y
from equation 2 we have logx=b/2y
divide logy/logx=2ay-b/3b
- 11 years agoHelpfull: Yes(1) No(6)
- we get two equations
log x +3 log y = a------(1)
2 log x + log y = b-----(2)
solving these two eq we will get ,
log y = (2a-b)/5
log x = (3b-a)/5
hence log x/log y = (2a-b)/(3b-a)......is the answer - 10 years agoHelpfull: Yes(1) No(0)
- @ hema could u explain this
log(x^2y)=log x^2+log y ? - 11 years agoHelpfull: Yes(0) No(0)
- @ranjeet
log(xy)=log x+log y
similarly log(x^2y)=log x^2+log y ..it's (x)^2*y not (x)^2y
I hope you got the idea.
regards..
- 11 years agoHelpfull: Yes(0) No(1)
- log y/log x=(2a-b)/(3b-a)
- 8 years agoHelpfull: Yes(0) No(0)
- log(xy^3) = logx + log(y^3) = log(x) + 3 log(y) = a --------------- eq(1)
log(yx^2) = logy + log(x^2) = log(y) + 2 log(x) = b --------------- eq(2)
Multiply eq(2) by 3 and substract it by eq(1), we get
log(x) = (3b-a)/5
Multiply eq(1) by 2 and substract it by eq(2), we get
log(y) = (2a-b)/5
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log(y)/log(x) = (2a-b)/(3b-a) - 8 years agoHelpfull: Yes(0) No(0)
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