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)=

Algebra ->  Equations -> 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)=      Log On


   

Question 202721: Could you help me about this equation?
2%2F%282-log%283%2Cx%29%29=6%2F%283%2Blog%283%2Cx%29%29-3
This is my solution but I my answer is undefine!!!
assume: log%283%2Cx%29=z
SO I have: 2/(2-z)=6/(3+z)-3
make common denomirator : 2%2F%282-z%29=%286-9%2B3z%29%2F%283%2Bz%29
and then : 2%283%2Bz%29=%282-z%29.%286-9%2B3z%29 [ =(2-z).(-3+3z)]
so we have: 6%2B2z=-6%2B6z%2B3z-3z%5E2
thus: 6%2B2z%2B6-6z-3z%2B3z%5E2
therefore: 3z%5E2-7z%2B12=0
Delta= b^2-4ac=49-144 but it's less than Zero and it's undefine!!!!!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This much of your solution is good:
2%2F%282-log%283%2Cx%29%29=6%2F%283%2Blog%283%2Cx%29%29-3
assume: log%283%2Cx%29=z
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:
2%2F%282+%2B+%28-z%29%29=6%2F%283%2Bz%29+%2B+%28-3%29
Now when we get the common denominator on the right side we might avoid the error you had:
2%2F%282+%2B+%28-z%29%29=6%2F%283%2Bz%29+%2B+%28-3%29%2A%28%283+%2B+z%29%2F%283+%2B+z%29%29
2%2F%282+%2B+%28-z%29%29=6%2F%283%2Bz%29+%2B+%28%28-9%29+%2B+%28-3z%29%29%2F%283+%2B+z%29%29
2%2F%282+%2B+%28-z%29%29=%28%28-3%29+%2B+%28-3z%29%29%2F%283+%2B+z%29%29
We can solve this by cross-multiplying:
2%2A%283+%2B+z%29+=+%282+%2B+%28-z%29%29%2A%28%28-3%29+%2B+%28-3z%29%29
6+%2B+2z+=+-6+%2B+-6z+%2B+3z+%2B+3z%5E2
6+%2B+2z+=+3z%5E2+%2B+%28-3z%29+%2B+%28-6%29
Subtracting 2z and 6 from both sides we get:
0+=+3z%5E2+%2B+%28-5z%29+%2B+%28-12%29
This will factor:
0+=+%28z+-+3%29%283z+%2B+4%29
This gives solutions of z = 3 or z = -4/3.
Substituting back in for z we get:
log%283%2Cx%29+=+3 or log%283%2Cx%29+=+%28-4%29%2F3
Rewriting these in exponential form we get
x+=+3%5E3 or x+=+3%5E%28-4%2F3%29
x+=+27 or x+=+1%2F%283%5E%284%2F3%29%29
x+=+27 or x+=+1%2F%28root%283%2C+3%5E4%29%29
If we want rationalized denominators we need a perfect cube in the denominator of the second solution:
x+=+27 or
x+=+27 or x+=+%28root%283%2C+3%5E2%29%29%2F%28root%283%2C+3%5E6%29%29
x+=+27 or x+=+%28root%283%2C+3%5E2%29%29%2F%283%5E2%29
x+=+27 or x+=+%28root%283%2C+9%29%29%2F9+=+%281%2F9%29root%283%2C+9%29

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.)