You can put this solution on YOUR website!
This one is a bit tricky for a couple of reasons. First of all, the "-x" which is the argument of "log(-x)" tells us that x must be a negative number. Logarithms must have a positive argument so -x must be positive, making x itself negative. The second difficulty is that this equation is nearly in "quadratic form". Quadratic form equations can be difficult anyway and this one is more difficult since the equation is not yet in quadratic form.
"Regular" quadratic equations have an x squared term and usually an x term, too. Quadratic form terms have a some-expression-with-x squared term and usually a some-expression-with-x term, too. We have a "log(-x)" squared term. If we had a "log(-x)" term instead of the "log(x^2)" term we would have a quadratic form equation. If we could somehow rewrite "log(x^2)" as an expression with "log(-x)" we would have a quadratic form equation.
So the first thing I am going to do is rewrite this log in terms of -x. Since we can rewrite the equation as:
Then we can use a property of logarithms to move the exponent out in front:
or
The equation is now in quadratic form. If you've done some of these then you may know how to proceed. If you don't see how to proceed, the use of a temporary variable can be helpful. Set the temporary variable equal to the "some-expression-with-x". So:
Let q = log(-x)
Then
Substituting this into our equation we get:
This is clearly a quadratic equation. To solve it we want one side equal to zero. (Since I like positive coefficients in front of the squared term I am going to subtract the entire left side from both sides (and rearrange the terms so that they are in standard form):
Now we factor:
From the Zero Product Property:
q-3 = 0
Solving:
q = 3
Of course we are not interested in a solution for "q". So we substitute back in for the q (Remember, it was just a temporary variable):
log(-x) = 3
To solve for x we have a little more work to do. The next step is to rewrite this in exponential form:
which simplifies to:
-x = 1000
Dividing (or multiplying by) -1 we get:
x = -1000
Last of all we check. We must check to ensure that our "solution" results in valid arguments of logarithms. (Valid arguments are positive.) Use the original equation to check:
Checking x = -1000:
Simplifying:
At this point we can see that both arguments are valid. So x = -1000 passes the required check. Note that it was OK for x to be negative. It's the arguments that must be positive! (If you want, you can finish checking to see if -1000 actually makes the equation true. Hint: log(1000000) = 6 and log(1000) = 3.)
P.S. Once you have done a few of these quadratic form equations, you will start to "see" the quadratic nature of your equation and be able to solve it without using a temporary variable.
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