SOLUTION: log2 (y+2)-log2(y-2)=1

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Question 706224: log2 (y+2)-log2(y-2)=1
Found 2 solutions by nerdybill, jsmallt9:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log2 (y+2)-log2(y-2)=1
log2 (y+2)/(y-2)=1
(y+2)/(y-2)=2^1
(y+2)/(y-2)=2
(y+2)=2(y-2)
y+2=2y-4
2=y-4
6=y




Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C+%28y%2B2%29%29-log%282%2C%28y-2%29%29=1
Solving equations like this usually starts with transforming the equation into one of the following forms:
log(expression) = log(other-expression)
or
log(expression) = number

With the "non-log" term of 1 on the right side, we will have difficulty achieving the "all-log" first form. So we will aim for the second form. For the second form we want only one logarithm. So we need to find a way to combine the two logs we have into one.

The two logs we have are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and and the same arguments. Our logs have the same bases, 2, but they have different arguments, y+2 and y-2.)

Fortunately there is a second way to combine logarithmic terms. There are two properties of logarithms which give us a way to combine logs:
  • log%28a%2C+%28p%29%29%2Blog%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29-log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29
These properties require the same bases and coefficients of 1. Our logs fit both requirements. We will use the second property because it, like our logs, has a "-" between the two logs. Using that property on our equation we get:
log%282%2C+%28%28y%2B2%29%2F%28y-2%29%29%29=1
We now have the second form.

With the second form 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%5E1+=+%28y%2B2%29%2F%28y-2%29
which simplifies to:
2+=+%28y%2B2%29%2F%28y-2%29

Now we have an equation where the variables are no longer "buried" inside a logarithm. We can solve this equation. First we'll multiply both sides by y-2 to eliminate the fraction:
%28y-2%29%2A2+=+%28y-2%29%28%28y%2B2%29%2F%28y-2%29%29
which simplifies to:
2y-4+=+y%2B2
Subtracting y from each side:
y-4+=+2
Adding 4:
y+=+6

Last of all we check. This is not optional whenever you solve logarithmic equations or when you multiply both sides of an equation by an expression that might be zero. We have done both.

Use the original equation to check:
log%282%2C+%28y%2B2%29%29-log%282%2C%28y-2%29%29=1
Checking x = 6:
log%282%2C+%28%286%29%2B2%29%29-log%282%2C%28%286%29-2%29%29=1
Simplifying...
log%282%2C+%288%29%29-log%282%2C%284%29%29=1
Since 8 is 2 cubed the first log is 3. And since 4 is 2 squared the second log is 2:
3-2=1
1+=+1
Check!!
So y=6 is the only solution to the equation. (Note: If the check had failed we would have to reject the solution. And since it was the only "solution" there would be no solution to the equation.)