You can put this solution on YOUR website! ln(x+2)+ln(x)=ln(35)
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) = other-expression
or
log(expression) = log(other-expression)
The terms in your equation are all logarithmic. So the second form will be easier to achieve. All we need to do in find a way to combine the two logarithms on the left side into one.
The two logarithms on the left are not like terms so we cannot just add them. (Like logarithmic terms have logarithms of the same base and equal arguments. Your terms have the same base, e, but different arguments, x+2 and x.)
Fortunately there is another way to combine logarithmic terms. There are properties of logarithms which allow us to combine logarithms:
These properties require logarithms of the same base with coefficients of 1. The arguments can be anything. Your logarithms meet these requirements. Since your logarithms have a "+" between them we will use the proeprty with the "+", the first one:
ln((x+2)*x)=ln35
which simplifies to
We now have the equation in the second form. With this form the next step is some simple logic. The equation tells us that one base e logarithm is equal to another base e logarithm. The only way for two base e logarithms to be equal is for the arguments to be equal. So:
This is now an equation we can solve. It is a quadratic equation so we want one side to be zero. Suntracting 35 from each side we get:
Now we factor (or use the Quadratic Formula). This factors easily:
(x+7)(x-5) = 0
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+7 = 0 or x-5 = 0
Solving these we get:
x = -7 or x = 5
You must check your answers on problems like this. You must ensure that the solutions make all arguments (and bases) of all logarithms positive. Any "solution" that makes any argument (or base) zero or negative must be rejected. And these rejected solutions can happen even if no mistakes have been made. This is why you must check.
When checking always use the original equation:
ln(x+2)+ln(x)=ln(35)
Checking x = -7
ln((-7)+2)+ln((-7))=ln(35)
We can already see that when x = -7 the arguments are going to be negative. So we reject this solution.
Checking x = 5
ln((5)+2)+ln((5))=ln(35)
We can already see that when x = 5 the arguments are going to be positive. So we have no reason to reject this solution. This is the required part of the check. The rest of the check will tell us if we made a mistake. You are welcome to finish the check.
So there is only one solution to your equation: x = 5.
Important: We rejected x = -7 not because the x was negative but because -7 made the arguments negative. If the equation had been
ln(-(x+2))+ln(-x)=ln(35)
we would have kept x = -7 and rejected x = 5!
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