SOLUTION: {{{log(2,x-6)}}} +{{{log(2,x-4)}}} -{{{ log( 2, x ) }}}=2 I condensed it and put the x as demoninator but now I dont know what to do with it.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: {{{log(2,x-6)}}} +{{{log(2,x-4)}}} -{{{ log( 2, x ) }}}=2 I condensed it and put the x as demoninator but now I dont know what to do with it.      Log On


   

Question 681864: log%282%2Cx-6%29 +log%282%2Cx-4%29 -+log%28+2%2C+x+%29+=2
I condensed it and put the x as demoninator but now I dont know what to do with it.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28x-6%29%29+%2Blog%282%2C%28x-4%29%29+-+log%28+2%2C+%28x%29+%29+=2
By "condense it and put x as the denominator" I hope you mean...
Using log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29 to combine the first two logs:
log%282%2C%28%28x-6%29%28x-4%29%29%29+-+log%28+2%2C+%28x%29+%29+=2
which simplifies to:
log%282%2C%28x%5E2-10x%2B24%29%29+-+log%28+2%2C+%28x%29+%29+=2
Using log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 to combine the remaining logs:
log%282%2C%28%28x%5E2-10x%2B24%29%2Fx%29%29+=2

The next step is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+k is equivalent to a%5Ek+=+p. Using this pattern on our equation we get:
2%5E2+=+%28x%5E2-10x%2B24%29%2Fx
which simplifies to:
4+=+%28x%5E2-10x%2B24%29%2Fx

Now that the logs are gone we can use "regular" algebra to solve. First let's multiply by x to eliminate the fraction:
4x+=+x%5E2-10x%2B24
Since this is a quadratic equation we want one side to be you. Subtracting 4x we get:
0+=+x%5E2-14x%2B24
Next we factor or use the Quadratic Equation. This factors fairly easily:
0+=+%28x-12%29%28x-2%29
From the Zero Product Property:
x+-+12+=+0 or x+-+2+=+0
Solving these we get:
x+=+12 or x+=+2

Last of all, we check our answers. This is not optional when solving these kinds of equations. You must ensure that any "solution" will make all arguments to all logarithms positive. If a "solution" makes any argument zero or negative then we must reject that solution. Use the original equation to check:
log%282%2C%28x-6%29%29+%2Blog%282%2C%28x-4%29%29+-+log%28+2%2C+%28x%29+%29+=2
Checking x = 12:

We can already see that all three arguments will be positive when x = 12. So this solution passes the required part of the check.

Checking x = 2:

We can see that the first two arguments will be negative when x = 2. So this solution fails the check and we reject it.

So there is just one solution to the equation: x = 12