You can put this solution on YOUR website! I assume these are base 2 logarithms. With the way you posted the problem is it difficult to tell, for example, if that first log is or log(2x). Please use language like:
"2 times the base 2 log of x ..." or "2 times the base 10 log of 2x ..." to make it clear. Tutors are more likelyto help when the problem is clear.
If I am incorrect about these being base 2 logarithms I hope you can figure out the correct solution after looking at the solution for base 2 logarithms.
Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = expression
or
log(expression) = log(expression)
Since your equation is has the "non-log" term of 2 on the right side, the first for will be easier to achieve. All we have to do is find a way to combine the two logarithms on the left into a single logarithm.
The terms on the left are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and same arguments.)
Fortunately there is another way to combine logarithmic terms. Two properties of logarithms:
These properties require the same base and coefficients of 1. Your logarithms have the same base but the first logarithm has a coefficient of 2. Fortunately there is a third property, , which allows us to move a coefficient into the argument as its exponent. Using this property on the first logarithm we get:
Now we can use the second property (because of the "-" between the logarithms) to combine them:
We finally have the first form. With this form the next step is to rewrite teh equation in exponential form. In general is equivalent to . Using this pattern on your equation we get:
which simplifies to:
To solve this let's first eliminate the fraction by multiplying both sides by (x+3):
which simplifies to:
Since this is a quadratic equation we want one side to be zero. And since the rest of the problem will be a little easier if the squared term has a positive coefficient, I am going to subtract 4x and 12 from each side:
Now we factor (or use the Quadratic Formula). This factors fairly easily:
(x-6)(x+2) = 0
From the Zero Product Property we know that one of the factors must be zero. So:
x-6 = 0 or x+2 = 0
Solving these we get:
x = 6 or x = -2
Solving logarithmic equations like yours requires that answers be checked to make sure that all arguments of logarithms are positive. If any arguments are found to be zero or negative, then those "solutions" must be rejected. (These rejected solutions can occur even if no mistakes were made in finding them! This is why even expert mathematicians must check for them.)
Use the original equation to check:
Checking x = 6:
We can see already that both arguments will be positive when x = 6. So there is no reason to reject this solution. The reset of the check will just tell us if we made a mistake. You are welcome to finish the check.
Checking x = -2:
We can see that the first argument is negative when x = -2. So we must reject this "solution". Rejecting this solution does not mean we made a mistake. It is just something that can happen any time you solve these kinds of equations.
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