Question 377488: log_7(x+6)-log_7(x+2)=log_7x
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! I'm guessing that "log_7" means "base 7 log". In the future, to make things clear, you might want to use(base 7 log(x+6))-(base 7 log(x+2)) = base 7 log(x). Or if you want to get really slick, type 3 left braces , "{", followed by "log(7, (x+6)) - log(7, (x+2)) = log(7, (x))" followed by 3 right braces, "}". When Algebra.com's software sees this it will show up as:
Tutors are more likely to help with problems that are clear.
Now to solve
When an equation has the variable in the argument of one or more logarithms, often the first step in solving the equation is transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)
Since your equation has no "non-log" terms the first form will be more difficult to achieve. So we will aim for the second form. This will require that we somehow combine the two logarithms into one on the left side. Fortunately there is a property of logarithms,  , which allows us to combine two logarithms with the same base, with coefficients of 1 and with a minus sign between them (like yours). Using this property on the left side we get:
We now have the second form. With this form the next step is to set the arguments equal. We can do this because each of the two logarithms represent exponents for 7. On the left the logarithm is the exponent for 7 that results in  and on the right side we have the exponent for 7 that results in x. The equation tells us that these two exponents are equal. And if you put equal exponents on 7's you should end up with equal results. So:
The logarithms are gone and we can now start using "regular" Algebra to solve for x. First we'll eliminate the fraction by multiplying both sides by (x+2):
which simplifies to:
This is a quadratic equation so we want one side to be zero. Subtractin x and 6 from each side we get:
Next we factor (or use the Quadratic Formula). This factors easily:
0 = (x+3)(x-2)
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So:
x+3 = 0 or x-2 = 0
Solving these we get:
x = -3 or x = 2
We must check our answers. Whenever you solve a logarithmic equation you must ensure that all arguments and bases of logarithms remain positive. And whenever you multiply both sides of an equation by an expression that might be zero, like we did when we multiplied both sides by (x+2), you must check make sure your answers do not make this expression zero. So we have two reasons (only one would be enough) for us to check our answers. When you check your answers, use the original equation:
Checking x = -3:
As we can already see, the last two logarithms have negative arguments when x = -3. So we must reject this "solution". (We would reject a solution if even only one argument (or base) was not positive.) Important: Rejecting this solution does not mean we made a mistake somewhere. This can happen any time you solve a logarithmic equation. This is why you must check you answers with equations like this!
Checking x = 2:
As we can already see, all of the logarithms will have positive arguments and bases. And if x=2 then the (x+2), that we used to multiply both sides of the equation, will be 4, not zero. So this solution looks good. The rest of the check will tell us if we made a mistake somewhere. If you want to finish the check you are welcome to do so. (Hint: You will need to use the property of logarithms again to combine the two logarithms on the left side.)
So the only solution to your equation is x = 2.
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