SOLUTION: Solve for all values of x: log4 (x^2+3x)-log4 (x+5)=1

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Question 404439: Solve for all values of x: log4 (x^2+3x)-log4 (x+5)=1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%284%2C+%28x%5E2%2B3x%29%29-log%284%2C+%28x%2B5%29%29=1
To solve equations where the variable is in the argument (or base) of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has a non-log term, the 1, it will be harder to reach the second, all log, form. All we need to do to reach the first form is to combine, somehow, the two logarithms into one.

The two logarithms are not like terms so we cannot subtract them. But fortunately there is a property of logarithms, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 which allows us to combine two logarithms as long as:
  • the bases are the same; and
  • the coefficients are 1's; and
  • there is a "-" between them.

Your logarithms meet all three requirements. So using this property to combine them we get:
log%284%2C+%28%28x%5E2%2B3x%29%2F%28x%2B5%29%29%29=1
We now have the first form. The next step is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq. Using this pattern on your equation we get:
%28x%5E2%2B3x%29%2F%28x%2B5%29=4%5E1
or
%28x%5E2%2B3x%29%2F%28x%2B5%29=4
This is an equation we can now solve. First we can eliminate the fraction by multiplying by (x+5):
%28%28x%5E2%2B3x%29%2F%28x%2B5%29%29%28x%2B5%29=%284%29%28x%2B5%29
x%5E2%2B3x=4x%2B20
This is a quadratic equation so we want one side to be zero. Subtracting 4x and 20 from each side we get:
x%5E2-x-20+=+0
Now we factor (or use the Quadratic Formula). This factors easily:
(x-5)(x+4) = 20
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-5 = 0 or x+4 = 0
Solving these we get:
x = 5 or x = -4

When solving these equations you must check your answer(s)! You must make sure that each solution makes all arguments (and bases) of logarithms positive. Any "solution" that makes an argument (or base) of a logarithm zero or negative must be rejected.

Always use the original equation to check:
log%284%2C+%28x%5E2%2B3x%29%29-log%284%2C+%28x%2B5%29%29=1
Checking x = 5:
log%284%2C+%28%285%29%5E2%2B3%285%29%29%29-log%284%2C+%28%285%29%2B5%29%29=1
which simplifies as follows:
log%284%2C+%2825%2B3%285%29%29%29-log%284%2C+%2810%29%29=1
log%284%2C+%2825%2B15%29%29-log%284%2C+%2810%29%29=1
log%284%2C+%2840%29%29-log%284%2C+%2810%29%29=1
We can see that all arguments and bases are positive. So there is no reason to reject this solution. This is the required part of the check. The rest of the check will just tell us if we made a mistake. You are welcome to finish the check if you want.

Checking x = -4:
log%284%2C+%28%28-4%29%5E2%2B3%28-4%29%29%29-log%284%2C+%28%28-4%29%2B5%29%29=1
which simplifies as follows:
log%284%2C+%2816%2B3%28-4%29%29%29-log%284%2C+%281%29%29=1
log%284%2C+%2816%2B%28-12%29%29%29-log%284%2C+%281%29%29=1
log%284%2C+%284%29%29-log%284%2C+%281%29%29=1
We can see that all arguments and bases are positive, even though x itself was negative. So there is no reason to reject this solution. This is the required part of the check. You are welcome to finish the check.

So there are two solutions to your equation: x = 5 or x = -4