SOLUTION: solve for x : ln(2x^2+4x)=5+ln2x

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Question 417249: solve for x :
ln(2x^2+4x)=5+ln2x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
ln%282x%5E2%2B4x%29=5%2Bln%282x%29
Solving equations where the variable is in the argument of a logarithm, like this equation, usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

With the "non-log" term of 5 in your equation, the first form will be easier to reach. So we want just a single logarithm. To get a single logarithm we will start by gathering both logarithms on the same side of the equation. Subtracting ln(2x) from each side we get:
ln%282x%5E2%2B4x%29+-+ln%282x%29=5
Now we have to do is find a way to combine the two logarithms into one.

The terms on the left side of the equation are not like terms so we cannot just subtract to combine them. (Like logarithmic terms have logarithms of the same bases and same arguments.)

There are two properties of logarithms which allow you to combine two into one:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29

These properties require logarithms of the same base with coefficients of 1. The two logs on the left meet these requirements. So we can use one of these properties to combine them. Since your logarithms have a "-" between them we will use the second proeprty (which also has a "-".) Using the second property on your equation we get:
ln%28%282x%5E2%2B4x%29%2F%282x%29%29+=+5
The fraction will reduce:
ln%28%282x%28x%2B2%29%29%2F%282x%29%29+=+5
ln%28%28cross%282x%29%28x%2B2%29%29%2Fcross%28%282x%29%29%29+=+5
ln%28x%2B2%29+=+5
We now have the equation in the desired first form. With this 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 (and the fact that the base of ln is "e") we get:
x%2B2+=+e%5E5
With the logarithm gone we can now solve for x. Subtracting 2 from each side we get:
x+=+e%5E5-2

Answers to equations like yours must be checked. You must ensure the answers make all arguments to all logarithms positive. Any "solution" that makes an argument zero or negative must be rejected (because arguments to all logarithms must be positive). These rejected solutions can happen any time and when they do happen it is not a sign that a mistake was made! So even expert mathematicians must check their answers on these equations.

Use the original equation to check:
ln%282x%5E2%2B4x%29=5%2Bln%282x%29
Checking x+=+e%5E5-2:
ln%282%28e%5E5-2%29%5E2%2B4%28e%5E5-2%29%29=5%2Bln%282%28e%5E5-2%29%29
With a little thought we can determine if these arguments are going to be positive. "e" is approximately 2.7. So e%5E5 will be greater than 2. And if e%5E5 is greater than 2 then e%5E5-2 will be positive. Since the arguments have only multiplication and addition with positive numbers, then all the arguments must work out to be positive numbers. So there is no reason to reject this solution. This is the required part of the check. The rest of the check is optional and will tell us if we made a mistake. You are welcome to finish the check.

So the only solution to your equation is x = x+=+e%5E5-2. This an exact expression for the solution to your equation. If you want a decimal approximation, replace "e" with a decimal approximation, raise it to the 5th power and then subtract 2. If your calculator does not have a button for "e", then either use
  • ln-1(1) (a calculator with a button for "ln" should also have a button for ln-1); or
  • 2.7182818284590451 (or a rounded off version of this).