Question 626338: factor and solve (lnx)^2 +1/2ln(3x^2)=20+ 1/2ln3
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
The clue to solving this is in the phrase "factor and solve". We use factoring to solve quadratic (and higher degree) equations.
So we need to look at this equation as a quadratic equation. The trick here is to realize that it is not a "pure" quadratic with an  term. The equation is in what is called "quadratic form". A quadratic form equation is an equation where a variable expression is raised to two different powers, one power being exactly twice the other power. (You'll see this shortly.)
In your equation we have ln(x) begin squared. And, if we know our properties of logarithms well, we can see that  can be rewritten in terms of ln(x) (to the first power). SO our equation is in quadratic form with an ln(x) being raised to the first and second powers. To solve it we must first rewrite . First we use the property of logarithms,  , to separate the 3 and the :
(Note how I used parentheses. This is critical when substituting an expression of two terms for an expression of one term.) Next we can use another property of logarithms,  , to move the exponent of the argument of  out in front:
We now have the equation in terms of ln(x) and  .
Until you've had some practice, quadratic form equations can be difficult to see how to solve. It can help to use a temporary variable. Let your temporary variable represent the variable expression of the lower power and then that variable squared will the the variable expression of twice that power:
Let t = ln(x).
Then 
Substituting these into our equation we get:
Now the equation should look more like a quadratic equation. To solve ti we start with simplifying, as usual. Using the Distributive property we can multiply on the left side:
which simplifies to:
Next we want the equation in  form. So we'll subtract the entire right side of the equation from both sides. (Fortunately the (1/2)ln(3)'s cancel out!)
Now we factor (or use the quadratic formula). This factors easily:
From the Zero Product Property:
t + 5 = 0 or t - 4 = 0
Solving these we get:
t = -5 or t = 4
Of course "t" is a temporary variable we just made up. What we really want are values for x. So at this point we substitute ln(x) back in for the "t":
ln(x) = -5 or ln(x) = 4
To solve these for x we rewrite them in exponential form. In general  is equivalent to  . Using this pattern (and the fact that "e" is the base of ln) on our equations we get:
 or 
Solutions to logarithmic equations should be checked. You must ensure that all arguments of logarithms remain positive. Use the original equation to check:
Checking  :
Simplifying:
Remembering that negative exponents mean reciprocals, not that you get a negative result, "e" to these negative powers will be positive. So we can see that the arguments are going to be positive numbers when . This is the required part of our check.
Checking  :
Simplifying:
Again we can see that the arguments of the logarithms are gping to be positive when . So this solution checks out, too.
So the solution to your equation is:
or
P.S.- When we checked a solution, if an argument of any logarithm had become zero or negative, we would reject that solution (even if it was the only solution we had found!)
- Eventually you will not need a temporary variable for these quadratic form equations. You will develop an "eye" and see how to go directly from
to
to
to
 or 
etc.
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