Problem 11 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{\frac{1}{10}} x=-3\)

Step-by-Step Solution

Answer

The solution is \( x = 1000 \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_{\frac{1}{10}} x = -3 \). This tells us that the logarithm with base \( \frac{1}{10} \) of \( x \) is equal to \( -3 \). We need to convert this logarithmic equation into an exponential form to solve for \( x \). Logarithmic and exponential forms are related as \( \log_b a = c \) is equivalent to \( a = b^c \).

Step 2: Convert to Exponential Form

Convert the logarithmic equation \( \log_{\frac{1}{10}} x = -3 \) to its equivalent exponential form. According to the logarithmic identity, \( x \) can be expressed as: \[ x = \left( \frac{1}{10} \right)^{-3} \]

Step 3: Simplify the Exponential Expression

Evaluate \( \left( \frac{1}{10} \right)^{-3} \). When a fraction is raised to a negative power, it involves taking the reciprocal of the fraction and raising it to the positive power. Hence: \[ \left( \frac{1}{10} \right)^{-3} = \left( 10 \right)^3 \]Now calculate \( 10^3 \): \[ 10 \times 10 \times 10 = 1000 \] Therefore, \( x = 1000 \).

Step 4: Verify the Solution

To verify, plug \( x = 1000 \) back into the original equation and see if it satisfies the equation:\[ \log_{\frac{1}{10}} 1000 \]Convert 1000 into a power of 10: \( 1000 = 10^3 \). Then take the reciprocal as the base is \( \frac{1}{10} \) which is \( 10^{-1} \). Now, recall that:\[ \log_{10^{-1}} 10^3 = \log_{10^{-1}} (10^{-1})^{-3} = -3 \] Hence, the solution is verified.

Key Concepts

Exponential Form

Exponential form is a fundamental concept in mathematics that helps us solve logarithmic equations. The relation between logarithms and exponents is crucial here. When we have an equation like \( \log_b a = c \), it means that \( b \) (the base) raised to the power of \( c \) gives \( a \).
This equivalence allows us to convert complex logarithmic problems into simpler exponential forms, which are often easier to work with.Consider the equation \( \log_{\frac{1}{10}} x = -3 \). By converting this into exponential form, we reframe it to \( x = \left( \frac{1}{10} \right)^{-3} \).
This way, we use our knowledge of how exponents operate on fractions to find a solution. Once in the exponential form, calculating the value becomes straightforward with basic arithmetic skills. This conversion technique makes tackling logarithmic equations intuitive and less daunting.

Negative Exponents

Negative exponents can initially seem confusing, but understanding them can greatly simplify complex equations. A negative exponent in an expression like \( a^{-n} \) prompts us to take the reciprocal of \( a \) and then raise it to the positive exponent \( n \).
For example, if we start with \( \left( \frac{1}{10} \right)^{-3} \), we flip the fraction, resulting in \( 10^3 \).
This is because the operation essentially 'undoes' the division implied by the \( \frac{1}{10} \).This principle is particularly helpful because it turns potentially complex operations into simple multiplication or division. By understanding and applying the rule of negative exponents, even challenging expressions become manageable, allowing for easier evaluation and problem-solving in mathematical contexts.

Logarithm Base Change

Sometimes in mathematics, we need to change the base of a logarithm to make an equation simpler or more solvable. This process is known as base change or base conversion. Let's explore the mechanics of how it works.
Suppose we have \( \log_b a \) and we want to change the base to a different base \( k \). We can use the formula:

• \( \log_b a = \frac{\log_k a}{\log_k b} \)

This formula provides a way to easily change the base of a logarithm, making it more adaptable to various equations or conditions.
For instance, if the base is a fraction, like in our case with \( \log_{\frac{1}{10}} \), the property simplifies evaluating the expression by switching to a base that might be easier to manage, such as ten. This method not only provides more flexibility in solving problems but also strengthens understanding of how logarithms are manipulated across different bases in mathematics.