SOLUTION: Could you help me about this equation? {{{2/(2-log(3,x))=6/(3+log(3,x))-3}}} This is my solution but I my answer is undefine!!! assume: {{{log(3,x)=z}}} SO I have: 2/(2-z)=

Question 202721: Could you help me about this equation?

This is my solution but I my answer is undefine!!!
assume:
SO I have: 2/(2-z)=6/(3+z)-3
make common denomirator :
and then : [ =(2-z).(-3+3z)]
so we have:
thus:
therefore:
Delta= b^2-4ac=49-144 but it's less than Zero and it's undefine!!!!!

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
This much of your solution is good:

assume:
giving: 2/(2-z)=6/(3+z)-3

Then we get a little off track finding the common denominator and subtracting. Since subtractions can cause so many problems I strongly encourage you to change subtractions to additions. So I would rewrite this as:

Now when we get the common denominator on the right side we might avoid the error you had:

We can solve this by cross-multiplying:

Subtracting 2z and 6 from both sides we get:

This will factor:

This gives solutions of z = 3 or z = -4/3.
Substituting back in for z we get:
or
Rewriting these in exponential form we get
or
or
or
If we want rationalized denominators we need a perfect cube in the denominator of the second solution:
or
or
or
or

Both of these answers check. (One should always check because we need to avoid extraneous solutions like those that would make the argument of a log function negative, make the radicand of an even-numbered root negative, denominators zero, etc.)
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