Problem 3 - Thomas Calculus

Question

Simplify the expressions. a. \(2^{\log _{4} x} \) b. \(9^{\log _{3} x}\) c. \(\log _{2}\left(e^{(\ln 2)(\sin x)}\right)\)

Step-by-Step Solution

Answer

a. \(\sqrt{x}\); b. \(x^2\); c. \(\sin x\).

Step 1: Understanding the Expression a

The expression is written as \(2^{\log_{4} x}\). We need to simplify it by recognizing the relationship between the base of the exponent and the logarithm base.

Step 2: Rewrite Using Change of Base Formula for a

Change \(\log_{4} x\) to base 2 using the change of base formula: \(\log_{4} x = \frac{\log_{2} x}{\log_{2} 4}\). Here, \(\log_{2} 4\) is 2, so the expression becomes \(2^{\frac{\log_{2} x}{2}}\).

Step 3: Simplification for a

Now, we write the expression \(2^{\frac{\log_{2} x}{2}}\) as \((2^{\frac{1}{2}})^{\log_{2} x}\). Given that \(2^{\frac{1}{2}} = \sqrt{2}\), this simplifies to \(x^{\log_{2} \sqrt{2}} = x^{\frac{1}{2}} = \sqrt{x}\).

Step 4: Understanding the Expression b

The expression is \(9^{\log_{3} x}\). Recognize the relationship between base 9 and base 3.

Step 5: Rewrite Base for b

Rewrite 9 as \(3^2\), so the expression becomes \((3^2)^{\log_{3} x}\).

Step 6: Simplification for b

Using the exponent rule \((a^m)^n = a^{mn}\), rewrite it as \(3^{2 \times \log_{3} x}\). Simplifying gives \((3^{\log_{3} x})^2 = x^2\).

Step 7: Understanding the Expression c

The expression is \(\log_{2}\left(e^{(\ln 2)(\sin x)}\right)\). Simplify it using logarithmic and exponential properties.

Step 8: Use Logarithmic Property for c

Apply the property \(\log_{b}(a^c) = c \log_{b}(a)\). The expression becomes \((\ln 2)(\sin x) \log_{2} e\).

Step 9: Simplification for c

Given that \(\log_{2} e\) can be rewritten using the change of base formula as \(\frac{\ln e}{\ln 2} = \frac{1}{\ln 2}\), substitute back to get \(\sin x\).

Key Concepts

Change of Base Formula

The change of base formula is often used to simplify logarithmic expressions when the bases are inconvenient. It states:

• \( \log_{b} a = \frac{\log_{k} a}{\log_{k} b} \)

This means you can take the logarithm with any base \(k\) and divide it by the logarithm of \(b\) with the same base \(k\).
You might most often see values like base 10 or base \(e\) used for \(k\). This is because they simplify to \(\log\) and \(\ln\) respectively, which are common in many mathematical problems.
In the original exercise example, applying the change of base formula allows for simplifying the expression \(2^{\log_{4} x}\). By changing \(\log_{4} x\) to \(\log_{2} x\), the base becomes simpler to work with as they share a relationship (\(4 = 2^2\)). Further simplification follows by realizing these similarities.

Exponent Rules

Exponent rules are critical in simplifying expressions that involve powers and roots. Here are some of the key rules you might use:

• \((a^m)^n = a^{m \times n}\)
• \(a^m \times a^n = a^{m+n}\)
• \(a^{-n} = \frac{1}{a^n}\)
• \(a^0 = 1\)

In the provided exercise, these rules help us simplify expressions like \((3^2)^{\log_{3} x}\). Applying the rule \((a^m)^n = a^{m \times n}\), the iteration \(3^{2 \times \log_{3} x}\) allows us to further simplify into \((3^{\log_{3} x})^2 = x^2\).
This is all thanks to understanding and applying these exponent rules, converting complicated expressions into simpler, more recognizable forms.

Logarithmic Properties

Logarithms bring a set of properties that simplify the processes of solving equations and expressions. Some major properties include:

• \(\log_{b}(a \times c) = \log_{b} a + \log_{b} c\)
• \(\log_{b}(a/c) = \log_{b} a - \log_{b} c\)
• \(\log_{b}(a^c) = c \log_{b} a\)
• \(\log_{b} b = 1\)
• \(\log_{b} 1 = 0\)

Using these properties makes expressions easier to interpret and reduce. For example, the property \(\log_{b}(a^c) = c \log_{b} a\) is applied in the solution to simplify \(\log_{2}(e^{(\ln 2)(\sin x)})\).
This expression is transformed using the property, yielding \((\ln 2)(\sin x) \log_{2} e\).
Further simplification recognizes \(\log_{2} e\) and its equivalent \(\frac{1}{\ln 2}\) via the change of base formula. Each step reduces complexity and reveals the beauty of logarithmic properties in handling what otherwise would be difficult calculations.