Question #ece0c

1 Answer
Aditya Banerjee.
Oct 5, 2017

Let's do it...

Explanation:

For solving this one, one should be aware of three formulas#rarr#

  1. #color(red)(logab=loga+logb)#.
  2. #color(blue)(alogb=logb^a)#.
  3. #color(green)(log10=1)#.

Now given, #logx+logy=4#
#:.log(xy)=4.log10#...from 1. & 3.
#:.log(xy)=log(10^4)#...from 2.
#:.color(red)(xy=10^4)#...Taking antilog in both sides.

Again given, #logx+2logy=3#.
#:.logx+log(y^2)=3.log10#...from 2. & 3.
#:.log(x.y^2)=log(10^3)#...from 1. & 2.
#:.color(red)(xy^2=10^3)#...Taking antilog in both sides.

Now you have got two simplified equations#rarr#

#color(red)(xy=10^4)#...(a).
#color(blue)(xy^2=10^3)#...(b).

Now, solve for the two equations:

#(b)/(a) rarr# #(xy^2)/(xy)=10^3/10^4#
So, #y=10^-1#

&, substituting value of #y# in #x rarr#
#x=10^4/10^-1=10^5#

Therefore: #color(red)(x=10^5)# & #color(blue)(y=10^-1)#

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