What is the Solution to the Logarithmic Discrepancy Equation?

[physicsdreams]

Homework Statement

log(x)-log(x+4)=2

Homework Equations

knowledge of logs

The Attempt at a Solution

This question was on a test, and I found the answer to be approximately -400/99.
I was marked off, because apparently the answer is no solution?
I checked wolframalpha, and they say the answer is -400/99

Am I right?

Thanks

 

[micromass]

You have defined the logarithm only for positive values of x. So log(-1) is not defined. The logarithm is a function that takes in positive values and spews out a real number.

It is possible to define the logarithm for negative numbers, but that's not what your teacher wants. Your teacher works with the real logarithm, and thus is not defined for negative numbers.

 

[ehild]

Sorry, you are not right. The logarithm of a negative number is not defined for real numbers. The original equation log(x)-log(x+4)=2 is equivalent with log(x/(x+4))=2 only for x>0. You have to exclude the impossible solutions before you start to solve an equation.

ehild

 

[physicsdreams]

You have to exclude the impossible solutions before you start to solve an equation.

ehild

Why must I exclude the "impossible equations" if they work? Can you please define impossible (graphically? algebraically)

Thanks

 

[physicsdreams]

Sorry, you are not right.

ehild

Here are the wolframalpha results for the equation:

http://www.wolframalpha.com/input/?i=log(x)-log(x+4)=2+solve+for+x&a=*FunClash.log-_*Log10.Log-

Why does it give me a correct solution?

Suppose the problem allowed for all possible answers (negative or possible) as long as they 'worked', would my answer then work?

Thanks

 

[D H]

Why does it give me a correct solution?

WolframAlpha is giving a solution in terms of the definition it uses for the function log₁₀(z). This is not the same as the definition you were given in your class / your text. The domain for the logarithm function as defined for your class is the positive real numbers. Plug your solution, x = -400/99, back into the original equation and you'll get log(-400/99)+log(-4/99)=2. Does that make any sense in terms of what you have been taught?

Sometimes when you solve a problem you will get superfluous answers. You need to learn to recognize that some answers are superfluous. A completely different example: Suppose you have calculated that the distance d between a pair of points is given by d²-d=2. This has two solutions, d=2 and d=-1. That second result (d=-1) is superfluous because distance can never negative. The only solution is d=2.Edit
I'll walk through the steps I assume you made to obtained an answer of -400/99.
Note: I am assuming that log(x) means the base 10 logarithm. (You wouldn't get -400/99 if log(x) indicates the natural log.)

Problem statement:
log(x)-log(x+4)=2

Step 1: Use the fact that log(a)-log(b)=log(a/b)
log(x/(x+4))=2

Step 2: Use the fact that log(a)=b is equivalent to a=10^b
x/(x+4)=100

Step 3: Multiply both sides by x+4
x=100(x+4)

Step 4: Solve for x
x=-400/99

Each step introduces the possibility of a superfluous solution. For example, the equivalent of step 3 is used in invalid but creative ways to prove that 1=2. In this case, it is that first step that is problematic. log(a)-log(b)=log(a/b) is valid only if log(a) and log(b) are defined. You should always check your work.

 

[physicsdreams]

Thanks D H.

But suppose that my teacher did defined the logarithm for all numbers both positive and negative. Is my answer then valid?

 

[ehild]

The (complex) logarithm function can be defined for complex numbers, among them for negative real ones. You need to know which logarithm is meant in the problem. But the variable of the complex logarithm is usually written as "z" instead of x. If you have not learned about complex numbers yet, your teacher could not define logarithm for negative numbers.

ehild