Problem 61 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The value of \( x \) is 10.

Step 1: Identify the Equation

The given problem is \( \log_{x} 0.01 = -2 \). This is a logarithmic equation where \( x \) is the base of the logarithm, \( 0.01 \) is the argument, and \( -2 \) is the result of the logarithm.

Step 2: Convert to Exponential Form

Using the definition of a logarithm, we can convert the equation into exponential form: \( x^{-2} = 0.01 \). This means that the base \( x \) raised to the power of \(-2\) equals \(0.01\).

Step 3: Solve for the Base

To find \( x \), take the reciprocal of \( x^{-2} \), which leads to \( x^{2} = \frac{1}{0.01} \). We know that \( 0.01 = 10^{-2} \), so \( \frac{1}{0.01} = 10^2 = 100 \). This gives us \( x^2 = 100 \).

Step 4: Find the Value of \( x \)

Take the square root of both sides of the equation \( x^2 = 100 \) to find \( x \). Thus, \( x = \sqrt{100} \). The principal square root yields \( x = 10 \), since \( x > 0 \) in logarithmic equations.

Key Concepts

Logarithmic Equations

A logarithmic equation involves expressing an exponent in terms of a logarithm. In the problem we are solving, \( \log_{x} 0.01 = -2 \), this equation shows how many times we need to multiply the base \( x \) to get to \( 0.01 \). It translates the expression of exponential growth or decay into a logarithmic function. It's important to remember that the equation structure is \( \log_{b}(y) = n \), where \( b \) is the base, \( y \) is the argument (the number we are evaluating the log of), and \( n \) is the result. By working with logarithmic equations, you can easily find unknown exponents.

Exponential Form

When we talk about converting a logarithmic equation to exponential form, we are essentially rearranging it back to its roots. Exponential notation describes repeated multiplication of a base. For our given equation, \( \log_{x} 0.01 = -2 \), the exponential form is \( x^{-2} = 0.01 \). This conversion implicitly rewires the logarithmic relationship back to its basic, intuitive concept of exponentiation, where the base \( x \) raised to the power \(-2\) yields \( 0.01 \). It is like reading the original equation from right to left: the base raised to an answer of \(-2\) gives us the number \( 0.01 \). Thus, converting logarithmic equations to exponential form is an essential skill that allows a clearer understanding and solution-finding process.

Base of a Logarithm

The base in a logarithmic function is the "big boss," if you will. It tells us what number we're repeatedly multiplying. In our specific problem, \( x \) is acting as this base. The base is crucial because it lays the foundation for the entire logarithmic relationship.
Understanding that the base \( x \) becomes \( x^{-2} = 0.01 \) in exponential form is key to solving the equation.

• The base dictates the logarithm’s behavior.
• Changing the base changes the entire problem's context.
• Knowing the base helps transition from the logarithmic form to the exponential form.

For our specific equation, solving for the base means finding the original value that when multiplied by itself minus its reciprocal power, gives the argument \( 0.01 \).

Square Root

Once we have the equation \( x^{2} = 100 \), taking the square root is our next step. Finding a square root involves determining which number multiplied by itself gives the original figure. In our case, we take \( \sqrt{100} \) yielding \( 10 \) because \( 10 \times 10 = 100 \).
The square root operation helps simplify and solve the exponential equations involved when converting from logarithmic form. In simpler terms, the square root helps undo the "squaring" process.

• The principal square root of a number points us to its positive root.
• In mathematics, when dealing with equations, we often take the principal square root to ensure positive solutions, especially in the context of logarithms.
• The square root of a perfect square is simple and direct, which is why in our problem, \( x = 10 \) is the seamless result.

Understanding how to correctly use the square root in this context leads to solving for \( x \) efficiently.