Question 275070: PLEASE HELP!!
Solve for x:
2log(x) = log(4) + log(x+3)
Solution Set: ______ ?
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
When solving logarithmic equations where the variable is in the argument or the base of the logarithm, you want to start by getting the equaiton into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)
Since your equation already has logarithms (and nothing but logarihtms) on both sides, we'll aim for the second form. This will require thatwe get rid of the 2 in front of the logarithm on the left- we combine the two logarithms on the right in to one logarithm
One the left side we can use one of the properties of logarithms,  , to move the 2 into the argument as an exponent:
On the right side we can use another property of logarithms,  , to combine the two logarithms into one:
which simplifies to:
We now have achieved the second form. With this form we have an equation that says the log of one expression equals the log of another expression. Since these are base 10 logarithms we can word it this way: The exponent for 10 that results in  equals the exponent for 10 that results in 4x+12. Since these exponents are equal it must mean that and 4x+12 are equal, too:
We now have an equation without logarithms to solve. Since this is a quadratic equation we'll start by getting one side equal to zero (by subtractin 4x and 12 from each side:
Now we factor (or use the Quadratic Formula). This factors fairly easily:
From the Zero Product Property we know that this product is zero only if one of the factors is zero. So:
 or 
Solving these we get:
 or 
When solving logarithmic equations it is important, not just a good idea, to check your answers. We must make sure no bases or arguments of a logarithm are zero or negative. And when checking, always use the original equation:
Checking x = -2:
As we can see, the argument of the first logarithm is negative. For this reason we must reject this solution.
Checking x = 6:
which simplifies to:
using the properties again we get:
or
 Check!
So the only solution to your equation is x = 6.
As we can see, the argument of the first logarithm is negative. For this reason we must reject this solution.
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