SOLUTION: I am having a bit of difficulty understanding some basic logarithmic equations, whilst i understand the more complicated ones. I am confused in the use of brackets in the follow

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Question 228085: I am having a bit of difficulty understanding some basic logarithmic equations, whilst i understand the more complicated ones.
I am confused in the use of brackets in the following question:
log(x-6) - log2 = log(x+2) - log3
When I attempted this question I expanded the brackets so that the equation was
logx - log6 = logx + log2 - log3
however when i tried to work this through, I couldnt work out the answer.
It would be great to receive some help as the textbook doesn't explain the use of brackets and i have an exam in a few days!
Thankyou very much for your time
zfitzpatrick
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%28x-6%29%29+-+log%28%282%29%29+=+log%28%28x%2B2%29%29+-+log%28%283%29%29
When I attempted this question I expanded the brackets so that the equation was


As you already found out, this is not correct. "log" is the name of the logarithm function. log%28%28x-6%29%29 stands for the result or output of the log function when the input is (x-6). log%28%28x-6%29%29 is not a multiplication and it is incorrect to use the Distributive Property on it like you have.

In order to solve equations where the variable is in the argument of a logarithm, you need to modify the equation, using the properties of logarithms and proper Algebra, into one of the following forms:
  • log(expression) = other-expression; or
    • log(expression) = log(other-expression)

So we need to condense the two logarithms on each side into one. In order to do this we need to understand the properties of logarithms:
  • y%2Alog%28a%2C+%28x%29%29+=+log%28a%2C+%28x%5Ey%29%29
  • log%28a%2C+%28x%29%29+%2B+log%28a%2C+%28y%29%29+=+log%28a%2C+%28x%2Ay%29%29
  • log%28a%2C+%28x%29%29+-+log%28a%2C+%28y%29%29+=+log%28a%2C+%28x%2Fy%29%29

The first one allows us to move a coefficient in front the log function into the argument as an exponent. The other two allow us to combine the addition or subtraction of logs into a single log. (Note that these two properties require no coefficients (IOW coefficients of 1) in front of the logs. We use the first property to get rid of coefficients, if any, before using these last two properties.)

In your problem, the logs have no coefficients so we will not need the first property. There are also no additions of logs so we will not need the second property. We will use the third property on both sides of your equation to combine the two logs into one:
log%28%28x-6%29%29+-+log%28%282%29%29+=+log%28%28x%2B2%29%29+-+log%28%283%29%29
log%28%28%28x-6%29%2F2%29%29+=+log%28%28%28x%2B2%29%2F3%29%29
The equation is now in one of the desired forms. A bit of logic will take us through the next step. If the log of (x-6)/3 is equal to the log of (x+2)/2, then (x-6)/3 must be equal to (x+2)/2. If two values are equal then their logs are equal (and vice versa). If this does not make sense to you then you can take the long route:
  1. Subtract one of the logs from each side. This makes the equation: log%28%28%28x-6%29%2F3%29%29+-+log%28%28%28x%2B2%29%2F2%29%29+=+0
  2. Use the third property on the left side creating a very complex fraction: log%28%28%28%28x-6%29%2F3%29%2F%28%28x%2B2%29%2F2%29%29%29+=+0
  3. Rewrite the equation in exponential form: %28%28x-6%29%2F3%29%2F%28%28x%2B2%29%2F2%29+=+10%5E0
  4. Solve this equation.

We'll take the short route and just say:
Since log%28%28%28x-6%29%2F2%29%29+=+log%28%28%28x%2B2%29%2F3%29%29
%28x-6%29%2F2+=+%28x%2B2%29%2F3
Now we'll solve this. There are several paths to solve this. I like to get rid of fractions by multiplying both sides of the equation by the Lowest Common Denominator (LCD) of all the fractions. The LCD between 2 and 3 is 6 so:
6%28%28x-6%29%2F2%29+=+6%28%28x%2B2%29%2F3%29
3%28x-6%29+=+2%28x%2B2%29
Now we simplify and solve:
3x+-+18+=+2x+%2B+2
Subtract 2x from each side:
x+-+18+=+2
Add 18 to each side:
x+=+20
(You can check this with your calculator. Just be aware that even calculators use approximations for logarithms so the two sides may not be exactly equal. If they are not equal they should be very close.)