Problem 67 - Algebra A Combined Function

Question

Solve. $$ \log _{x} \frac{1}{7}=\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{49} \).

Step 1: Understand the Logarithmic Equation

The equation given is \( \log_{x} \frac{1}{7} = \frac{1}{2} \). This is a logarithmic equation where the base \( x \) is unknown and the argument is \( \frac{1}{7} \). The equation says that the logarithm of \( \frac{1}{7} \) with base \( x \) is equal to \( \frac{1}{2} \).

Step 2: Convert to Exponential Form

Convert the logarithmic equation \( \log_{x} \frac{1}{7} = \frac{1}{2} \) into its exponential form. Use the property \( \log_{b}(y) = n \) implies \( b^n = y \). Therefore, \( x^{\frac{1}{2}} = \frac{1}{7} \).

Step 3: Solve for the Base \( x \)

To isolate \( x \), square both sides of the equation \( x^{\frac{1}{2}} = \frac{1}{7} \). Thus, \( (x^{\frac{1}{2}})^2 = \left(\frac{1}{7}\right)^2 \). Simplifying gives \( x = \frac{1}{49} \).

Step 4: Verify the Solution

Verify by substituting \( x = \frac{1}{49} \) back into the original logarithmic equation. The expression \( \log_{\left(\frac{1}{49}\right)} \frac{1}{7} = \frac{1}{2} \) should hold true. This satisfies as \( \left(\frac{1}{49}\right)^{\frac{1}{2}} = \frac{1}{7} \).

Key Concepts

Exponential Form

A logarithmic equation can often seem complex, but transforming it into exponential form can simplify the process significantly. In logarithmic terms, we say if the logarithm of a number with base \( b \) equals \( n \), then \( b^n = y \). Applying this concept to the equation \( \log_{x} \frac{1}{7} = \frac{1}{2} \) helps us move to exponential form. Here, the unknown base \( x \) is raised to the power of \( \frac{1}{2} \), which results in \( \frac{1}{7} \). Therefore, the equation in exponential form is \( x^{\frac{1}{2}} = \frac{1}{7} \).

• A logarithm tells you how many times to multiply the base by itself to reach the argument.
• Converting to exponential form helps in solving equations with unknown bases.

Base of Logarithm

The base of a logarithm is a critical component in resolving a logarithmic equation. In our case, it was unknown from the beginning. To find the base of logarithm from \( \log_{x} \frac{1}{7} = \frac{1}{2} \), we need to deduce what value of \( x \) satisfies the equation when transformed to its exponential form.

• The base indicates the number which is repeatedly multiplied.
• Understanding how the base works can simplify navigation through complex problems.

Finding the base \( x \) becomes straightforward once expressed as \( x^{\frac{1}{2}} = \frac{1}{7} \). Squaring both sides results in \( x = \frac{1}{49} \), revealing the required base.

Solving Logarithmic Equations

Solving logarithmic equations often entails converting them to exponential form to clearly see the relationship between the variables. In our example, after moving to exponential form, we determined \( x^{\frac{1}{2}} = \frac{1}{7} \). Solving for \( x \) involves:

• Isolating \( x \) by squaring both sides to get \( x = \left( \frac{1}{7} \right)^2 \).
• Calculating \( \frac{1}{7} \times \frac{1}{7} \), resulting in \( x = \frac{1}{49} \).

Breaking the problem down into smaller steps helps avoid errors and ensures a logical flow of solving the equation. Handling logarithmic equations requires understanding both logarithmic and exponential rules.

By practicing these transformations, you can quickly solve similar problems.

Verification of Solutions

After determining the solution \( x = \frac{1}{49} \), it's crucial to verify if it satisfies the original equation. Verification ensures accuracy and confirms that no mistakes were made during solving. By substituting \( x \) back into the original logarithmic equation, \( \log_{\left(\frac{1}{49}\right)} \frac{1}{7} = \frac{1}{2} \), we check if this holds true.

• Calculate \( \left(\frac{1}{49}\right)^{\frac{1}{2}} \), which should equal \( \frac{1}{7} \).
• If both sides of the equation balance, the solution is correct.

Verifying solutions is a vital step that strengthens your confidence in mathematical processes, ensuring both understanding and correctness of the intended mathematical application.