SOLUTION: log base 7 (x+6) - log base7 (x-3)= log base7 4

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Question 817713: log base 7 (x+6) - log base7 (x-3)= log base7 4
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%287%2C+%28x%2B6%29%29+-+log%287%2C+%28x-3%29%29=+log%287%2C+%284%29%29
Solving equations like this usually starts with using algebra and/or properties of logs to transform the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(other_expression)

We can reach the second form if we can either eliminate a log on the left or find a way to combine the two logs on the left. Fortunately there is a property of logs, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29, which allows us to combine them:
log%287%2C+%28%28x%2B6%29%2F%28x-3%29%29%29=+log%287%2C+%284%29%29
We now have the second form.

The next step is to eliminate the logs. With the second form, we just set the arguments equal. (The only way two logs (of the same base) can be equal is if their arguments are equal.)
%28x%2B6%29%2F%28x-3%29=+4

Now that the logs are gone we can solve for x. Multiplying both sides by x-3 (to eliminate the fraction):
x%2B6=+4x-12
Subtracting x:
6=+3x-12
Adding 12:
18 = 3x
Dividing by 3:
6 = x

Last we check. This is not optional! A check must be made that each solution will make the bases and arguments of any logarithms valid. (Valid bases are positive but not 1 and valid arguments are positive.) If a "solution" makes any base or argument invalid then it must be rejected.

Use the original equation to check:
log%287%2C+%28x%2B6%29%29+-+log%287%2C+%28x-3%29%29=+log%287%2C+%284%29%29
Checking x = 6:
log%287%2C+%28%286%29%2B6%29%29+-+log%287%2C+%28%286%29-3%29%29=+log%287%2C+%284%29%29
We should already see that this solution passes the required check. The bases are 7's (valid) and the arguments will turn out positive (valid) after the adding and subtracting is done.