SOLUTION: Solve for x in the equation log6(2x + 8) − log6(x + 1) = 1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve for x in the equation log6(2x + 8) − log6(x + 1) = 1      Log On


   

Question 689908: Solve for x in the equation log6(2x + 8) − log6(x + 1) = 1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2C+%282x+%2B+8%29%29+-+log%286%2C+%28x+%2B+1%29%29+=+1
Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(expression)

If we can find a way to combine the two logs into one the equation will be in the first form. The two logs are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and the same arguments. Your logs have the same base, e, but the arguments, x+5 and x-2, are different.) Fortunately there is a property of logarithms that allow us to combine two logs with a "-" between them: log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29. This property requires the same bases and coefficients of 1. Your logs meet both requirements. Using this property we get:
log%286%2C+%28%282x+%2B+8%29%2F%28x+%2B+1%29%29%29+=+1
We now have the equation in the first form.

The next step with this form is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to a%5Eq+=+p. Using this pattern on your equation:
6%5E1+=+%282x+%2B+8%29%2F%28x+%2B+1%29
which simplifies to:
6+=+%282x+%2B+8%29%2F%28x+%2B+1%29

Now that the variable is out of the logarithms we can solve for it. We'll start by multiplying each side by x+1 to eliminate the fraction:
6%28x+%2B+1%29+=+%28%282x+%2B+8%29%2F%28x+%2B+1%29%29%28x%2B1%29
which simplifies to:
6x+%2B+6+=+2x+%2B+8
Subtracting 2x:
4x+%2B+6+=+8
Subtracting 6:
4x+=+2
Dividing by 4:
x+=+1%2F2