 # Absolute Value and Logarithms

Absolute values often turn up unexpectedly in problems involving logarithms. That’s because you can’t take the log of a negative number.Let’s first review the definition of the logarithm function:Logx = y ⇔ by = x(The double arrow is a bi-conditional, which means that one side is true if and only if the other side is true).There are several restrictions on this definition when using real numbers (those not involving i = √-1 ).
1. b cannot equal 1
2. b cannot be negative
3. x cannot be less than or equal to zero.
If any of these conditions are not met, we risk introducing imaginary solutions or an infinite number of solutions. To prevent this from happening, we must follow these rules when working with real numbers, and to do so we often need to introduce the absolute value. Example 1. Expand the function f(x,y) = log(xy).We must always have a positive number inside the log function, and since x and y are being multiplied together, they must then always have the same sign. Therefore, the domain of our function is xy>0 which means that f(x,y) lies in Quadrants I and III on the x-y plane.To expand, we must follow the above rules, with emphasis in this case on Rule 3. To make sure that the values inside the log are positive, we introduce the absolute value.log(xy) = log|x| + log|y|If we had not included the absolute values, log(x) + log(y) would only be defined when x and y are both positive (Quadrant I). Since the domain is different than that of log(xy), they cannot be equivalent functions.Example 2. Simplify the function g(x,y) = log|x| + log|y|It is important to notice that the above solution is not true in the reverse direction. g(x,y) = log|x| +  log|y| is defined for all real, nonzero values of x and y. Therefore, to simplify this function, we must make sure the domain remains all real nonzero numbers. To do so, we use the absolute value. Simplifying,log|x| + log|y| = log|xy|Example 3. Evaluate the indefinite integral ∫ 1⁄x dxThis common integral can be found in many integral tables. It is equal to ln|x| + C. The absolute value is important because this is an indefinite integral, which means x might range through the entire real number line (There is a singularity at x=0, but log(0) is undefined too). We introduce the absolute value into the log to ensure that the antiderivative is defined everywhere the integral is.When working in the real numbers with log functions, remember to always follow the three rules listed above. For many expressions involving logs, it is often necessary to introduce absolute values to ensure that these rules are met.

# Absolute Value and Square Roots

Absolute values often show up in problems involving square roots. That’s because you can’t take the square root of a negative number without introducing imaginary numbers (those involving i = √-1 ).

Example 1: Simplify √x².

This problem looks deceptively simple. Many students would say the answer is x and move on. However, that is only true for positive values of x.

Try x = 2

√x² =  x

Left-hand side: √2² =  √4 =  2

Right-hand-side:  2

The equality holds for x = 2 (and actually for all values of x that are ≥  0).

Try x = -5.

√x² =  x

Left-hand side: √(-5)² = √25 = 5

Right-hand-side: -5

Since 5 ≠ -5, this equation does not hold for x = -5. Notice that when x < 0, the left hand side is actually equal to -x. Therefore, for all x < 0, √x² = -x

To summarize, for all x ≥ 0, √x² = x, and for all x < 0, √x² = -x.

We can write this as a piecewise function as follows:

$\sqrt{x^2} = \begin{cases} x, & \text{if$x≥0$} \\ -x, & \text{if$x \lt 0$} \end{cases}$

Our assignment was to simplify, and a piecewise function is surely not simpler than √x². However, this piece-wise function is actually the exact definition of absolute value. Therefore, we can write √x² = |x|, and our simplification is complete. This holds true, for any expression in the square root that has a even exponent. Even though x² is positive, it can be product of two negatives ex: (-5) (-5) or two positives (5)(5). Therefore, the absolute value ensures that both cases are addressed in the solution.

Example 2: Solve for x:  x²=z+9

x²=z+9

To solve for x, we first take the square root of both sides.

√x²=√z+9

As we discussed earlier, √x² = |x|, so

|x| = √z+9

The equation isn’t quite solved for x yet. To remove the absolute value, we write:

x = ±√z+9, and our work is done.

When working on problems involving square roots, remember to always check the positive and negative cases and be careful that you don’t miss the absolute value.