Problem 57 - Algebra 2

Question

Solve \(\log _{\sqrt{a}} 3=\log _{a} x\) for \(x\) and explain each step.

Step-by-Step Solution

Answer

The solution for \(x\) is 9.

Step 1: Understand the Problem

We need to solve the equation \(\log_{\sqrt{a}} 3 = \log_a x\). This means we are finding \(x\) such that the logarithm of 3 with base \(\sqrt{a}\) is equal to the logarithm of \(x\) with base \(a\).

Step 2: Use Change of Base Formula

We use the change of base formula \(\log_b c = \frac{\log_k c}{\log_k b}\) for a common logarithm base. Let's change both sides of the equation to base \(a\):\[\log_{\sqrt{a}} 3 = \frac{\log_a 3}{\log_a \sqrt{a}} \]And the right side remains as \(\log_a x\).

Step 3: Simplify \(\log_a \sqrt{a}\)

We simplify \(\log_a \sqrt{a}\). The square root of \(a\) is \(a^{1/2}\). Thus, \(\log_a \sqrt{a} = \log_a a^{1/2} = \frac{1}{2} \log_a a = \frac{1}{2} \times 1 = \frac{1}{2}\).

Step 4: Set the Simplified Expression Equal to \(\log_a x\)

Substitute \(\frac{1}{2}\) into the left side: \[\frac{\log_a 3}{1/2} = \log_a x\]Multiply both numerator and denominator by 2:\[2 \times \log_a 3 = \log_a x\]

Step 5: Solve for x

Since \(2 \times \log_a 3 = \log_a x)\), we can exponentiate both sides to remove the log base \(a\):\[a^{2 \log_a 3} = x\]Using the property \(a^{\log_a b} = b\), we simplify:\[a^{\log_a (3^2)} = x\]Therefore, \(x = 9\).

Step 6: Conclusion: State Final Answer

We have found that the value of \(x\) satisfying the given equation is \(x = 9\).

Key Concepts

Change of Base Formula

When working with logarithms, sometimes it is necessary to change the base of a logarithm to make it easier to work with or to compare it to another logarithm.
The change of base formula comes in handy in these situations.
This formula is expressed as \( \log_b c = \frac{\log_k c}{\log_k b} \), and it allows us to convert a logarithm to a different base, usually a common one like 10 or \( e \).In our original exercise, we needed to convert the logarithm \( \log_{\sqrt{a}} 3 \) to a base \( a \).

• By using the change of base formula, we rewritten it as \( \frac{\log_a 3}{\log_a \sqrt{a}} \).
• This transformation simplifies the problem and aligns the bases, making it easier to solve the equation.

Understanding how to change the base is crucial whenever you encounter logarithms that do not share a common base, or when simplifying complex logarithmic expressions.

Exponentiation

Exponentiation is a mathematical operation involving a base and an exponent.
The exponent indicates how many times the base is multiplied by itself.
This concept is tied closely to logarithms, as they are essentially inverses of exponentiation.In the context of solving the equation \( \log_{\sqrt{a}} 3 = \log_a x \), exponentiation was used to eliminate the logarithm and solve for \( x \).

• We set \( 2 \times \log_a 3 = \log_a x \) and used the property of exponentiation that \( a^{\log_a b} = b \).
• This allowed us to rewrite the expression as \( a^{2 \log_a 3} = x \).

By then applying the rules of exponents, we further simplified to find \( x = 9 \).
Exponentiation simplifies operations with logarithms, enabling us to solve equations that initially seem complex.

Logarithmic Properties

Logarithmic properties are essential because they provide structure to mathematical expressions.
They help in simplifying equations and are useful tools for solving various mathematical problems.In the given problem, several logarithmic properties were applied:

• Product Rule: \( \log_b (mn) = \log_b m + \log_b n \). Although not directly used, understanding this rule is critical for manipulating logs.
• Power Rule: \( \log_b (m^n) = n \cdot \log_b m \). This was used when converting \( \log_a \sqrt{a} = \log_a a^{1/2} \) to \( \frac{1}{2} \log_a a \).

By simplifying \( \log_a a = 1 \), we applied these properties further, specifically:

• The power rule helped reduce the expression signifying that the logarithm of a base to its own power equals one times that power.

The ability to break down logarithms with these properties is essential for clear, accurate, and efficient problem-solving.