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Chapter 11: Problem 125

Show that \(\log _{b^{2}} x=\frac{1}{2} \log _{b} x\).

Short Answer

Expert verified
Yes, \(\log_{b^2} x = \frac{1}{2} \log_b x\) by using the change of base formula and simplifying.

Step by step solution

01

Understand the Problem

We need to show that the logarithmic expression \(\log_{b^2} x\) is equivalent to \(\frac{1}{2} \log_b x\). This involves using log properties to transform one side of the equation to the other.
02

Use the Change of Base Formula

Apply the change of base formula: \(\log_{b^2} x = \frac{\log_b x}{\log_b b^2}\). This formula allows us to express a logarithm with one base in terms of another.
03

Simplify the Logarithm in the Denominator

Calculate \(\log_b b^2\) using the power rule. The power rule states that \(\log_b b^2 = 2\log_b b = 2\), since \(\log_b b = 1\).
04

Substitute Back into the Equation

Substitute the value from Step 3 into the equation from Step 2: \(\log_{b^2} x = \frac{\log_b x}{2}\).
05

Recognize the Final Expression

The expression \(\frac{\log_b x}{2}\) is equivalent to \(\frac{1}{2} \log_b x\). Both have the same mathematical meaning.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Change of Base Formula
The change of base formula is a powerful way to convert a logarithm from one base to another. It's particularly useful when calculators can only compute logarithms for specific bases like 10 or 2. The formula states:
  • \( \log_{a} x = \frac{\log_{c} x}{\log_{c} a} \)
This means you can express \( \log_{a} x \) using another base \( c \).
In our exercise, we used base \( b \) to express \( \log_{b^2} x \) as \( \frac{\log_b x}{\log_b b^2} \).
It helps transform base \( b^2 \) into base \( b \), making calculations simpler and allowing us to apply other logarithm rules more easily. Understanding this formula is key in diverse math problems, especially when dealing with different bases.
Exploring the Power Rule for Logarithms
The power rule for logarithms is another handy tool in simplifying expressions. The rule states:
  • \( \log_b (m^n) = n \cdot \log_b m \)
This means that if you have a power inside the logarithm, you can bring the exponent out in front as a multiplier.
In the given problem, we used this rule to simplify \( \log_b b^2 \), turning it into \( 2 \cdot \log_b b \).
Since \( \log_b b = 1 \), it further simplifies to 2.
This step was crucial to connect the original expression \( \log_{b^2} x \) to \( \frac{1}{2} \log_b x \). Knowing how to apply the power rule allows you to tackle more complex logarithmic expressions with ease.
Demystifying Logarithmic Expressions
Logarithmic expressions can often look intimidating, but they follow simple patterns that make them manageable. A logarithm, in essence, answers the question: "To what power must the base be raised to produce a given number?"
  • Example: \( \log_b x \) asks for the power to which \( b \) must be raised to result in \( x \).
In our exercise, we transformed \( \log_{b^2} x \) into a more manageable form using properties such as the change of base and power rule.
This shows the flexibility of logarithms and how they can be manipulated into simpler forms.
Overall, once you understand the basic rules logarithms follow, they become much less daunting and significantly more useful in solving mathematical problems.

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Most popular questions from this chapter

Solve each equation. $$x^{\log x}=10,000$$

a. \(\log 5 x=1.7\) b. \(\ln 5 x=1.7\)

Use a calculator to evaluate each expression, if possible. Express all answers to four decimal places. See Using Your Calculator: Evaluating Base-e (Natural) Logarithms. $$\ln 1.72$$

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest tenth. See Using Your Calculator: Solving Exponential Equations Graphically or Solving Logarithmic Equations Graphically. $$ \ln (2 x+5)-\ln 3=\ln (x-1) $$

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log _{5}(7+x)+\log _{5}(8-x)-\log _{5} 2=2 $$
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