the values of an expression

[Vali]

There are a,b real numbers which satisfy this \(\displaystyle 2lg(x-2y)=lg(x)+lg(y)\).
I need to find the values of this expression: \(\displaystyle \frac{x}{y}\)
My try: lg(x^2-4xy+4y^2)=lgxy
x^2-4xy+4y^2=xy / : y^2
(x/y)^2 -5(x/y)+4=0
I noted x/y=a and I got a^2-5a+4=0 so a=4 or a=1 so the values of x/y are 4 and 1 but the right answer is just 4.Why 1 is not good?

 

[pka]

There are a,b real numbers which satisfy this \(\displaystyle 2lg(x-2y)=lg(x)+lg(y)\).
I need to find the values of this expression: \(\displaystyle \frac{x}{y}\)
My try: lg(x^2-4xy+4y^2)=lgxy
x^2-4xy+4y^2=xy / : y^2
(x/y)^2 -5(x/y)+4=0
I noted x/y=a and I got a^2-5a+4=0 so a=4 or a=1 so the values of x/y are 4 and 1 but the right answer is just 4.Why 1 is not good?

Vali, please review what you posted. What in the world do \(\displaystyle a~\&~b\) have to do with any thing in the question?
What does \(\displaystyle lgx\) mean? Are you trying using to use \(\displaystyle \log(x)\)? If so the properties make no sense.
If not \(\displaystyle \log\) what is it?

 

[Harry_the_cat]

Yes lg seems to mean log. I see what you've done.
If x/y =1 then x=y.

If x and y are both greater than 0 (which they both must be for the RHS to be defined) then x-2y=x-2x=-x which is negative.

This would make the expression on the LHS undefined as you cant take the log of a neg number. So 1 is not a valid solution for the original equation.

For this reason when solving equations involving log you should always check your solutions fit the original equation.

Note also that if x/y = 4 then x =4y and the LHS will be defined so 4 is a valid solution.

 

[Steven G]

To add more to what the cat said WHENEVER you solve a log problem the answers you get are only possible answers! You MUST go back and confirm that these values work for the domain.

EX: 2logx = logx². You can then say that logx² = logx² and that x²=x² so that x can be any real number. But this not true as x must be positive (for 2logx to make sense!)

 

[Vali]

Thank you for your help I understood.
@pka my bad, there are x,y real numbers.