Problem 61 - Elementary and Intermediate Algebra

Question

Solve for \(x\). See Example 3 . $$ \log _{x} 0.01=-2 $$

Step-by-Step Solution

Answer

The solution is \(x = 10\).

Step 1: Rewriting the Logarithmic Equation

To solve the equation \(\log_{x} 0.01 = -2\), we start by rewriting the logarithmic equation in its exponential form. The logarithm \(\log_{x}(y) = b\) can be rewritten as \(x^b = y\). Therefore, \(-2\) is the exponent we raise \(x\) to in order to get \(0.01\). This results in the equation \(x^{-2} = 0.01\).

Step 2: Solving the Exponential Equation

The equation \(x^{-2} = 0.01\) can be rewritten to remove the negative exponent: \(\frac{1}{x^2} = 0.01\). To simplify, take the reciprocal of both sides, resulting in \(x^{2} = \frac{1}{0.01}\). Calculate \(\frac{1}{0.01}\) to get \(100\). Thus, we have \(x^2 = 100\).

Step 3: Finding the Solution for \(x\)

To solve for \(x\), take the square root of both sides of the equation \(x^2 = 100\). Remember that taking the square root yields two solutions, a positive and a negative: \(x = \pm \sqrt{100}\). Therefore, \(x = 10\) or \(x = -10\).

Step 4: Checking the Validity of Solutions

We need to check both potential solutions in the context of the base of a logarithm. A base must be positive and cannot be 1. Therefore, \(x\) must be greater than 0 and not equal to 1. Since \(x = 10\) is valid and positive, but \(x = -10\) is not a valid logarithmic base, we discard \(x = -10\). Thus, the only valid solution is \(x = 10\).

Key Concepts

Exponential Form

When solving logarithmic equations, one common technique is converting the logarithmic form into an exponential form. This is because exponential equations are often easier to manipulate and solve. To do this conversion, use the formula:

• If you have a logarithmic equation like \( \log_b(y) = x \), it can be rewritten in exponential form as \( b^x = y \).

In our exercise, the equation \( \log_{x} 0.01 = -2 \) can be transformed into the exponential expression \(x^{-2} = 0.01\). This tells us that raising \(x\) to the power of \(-2\) equals \(0.01\). By rewriting the problem this way, we have set a foundation that makes the following steps easier to carry out.Understanding this transformation can be extremely helpful when dealing with more complex logarithmic problems, as it lays out a simpler path by turning a log problem into a power problem.

Check Solutions

Checking solutions in any mathematical problem is crucial, especially when dealing with logarithmic problems that might yield multiple solutions. After solving the equation and finding potential solutions, it is important to verify which, if any, satisfy the conditions set by the problem.

• In the context of logarithmic equations, the base \(x\) must be positive and not equal to 1.
• This restriction arises because the logarithm base determines the behavior of the logarithmic function.

For our exercise, after solving, we found \(x = 10\) and \(x = -10\). However, since a logarithm cannot have a negative base and the base \(x\) cannot equal 1, we need to discard \(x = -10\). Thus, the only solution that satisfies this condition is \(x = 10\).This checking step ensures that all mathematical solutions are validated against the constraints, thus only keeping the sensible answer.

Square Root

Taking the square root is a common step in solving equations that involve exponents, particularly when you have equations like \(x^2 = a\). Square roots are useful because they allow us to "undo" squaring and make the variable more manageable. When you compute the square root of both sides of \(x^2 = a\), you must consider both the positive and negative roots. This is because both \((\sqrt{a})^2 = a\) and \((-\sqrt{a})^2 = a\) as well.

• In our specific example, we have \(x^2 = 100\).
• The square root of 100 is \(\pm 10\), so \(x = 10\) or \(x = -10\).

While both solutions are mathematically valid for a generic square, further steps (like the checking step) will limit which of these can be a base of a logarithm.

Negative Exponents

Negative exponents indicate division or a reciprocal relationship. For example, a negative exponent can be seen as an instruction to "flip" the base number and raise it to the positive equivalent of that exponent.

• If \(x^{-n}\) is given, then one can interpret it as \(\frac{1}{x^n}\).

This is particularly helpful when solving equations like \(x^{-2} = 0.01\). Converting a negative exponent into a division problem can make it more straightforward to work with:

• Rewrite as \(\frac{1}{x^2} = 0.01\).
• Then take reciprocal of both sides to find \(x^2 = 100\).

Understanding negative exponents is a cornerstone of algebraic manipulation, as it allows one to reframe what can initially appear to be complex fractions into a more manageable form.