Problem 15 - Algebra and Trigonometry

Question

\(7-18\) Evaluate the expression. $$ \log _{4} 16^{100} $$

Step-by-Step Solution

Answer

The expression evaluates to 200.

Step 1: Apply Exponent Rule for Logarithms

The expression inside the logarithm, \(16^{100}\), allows us to use the exponent rule for logarithms: \( \log_{b}(a^n) = n \cdot \log_{b}(a) \). Therefore, \( \log_{4}(16^{100}) \) can be rewritten as \( 100 \cdot \log_{4}(16) \).

Step 2: Simplify \(\log_{4}(16)\)

We need to calculate \( \log_{4}(16) \). Since \(16 = 4^2\), we find that \( \log_{4}(16) = \log_{4}(4^2) = 2 \cdot \log_{4}(4) \).

Step 3: Evaluate \(\log_{4}(4)\)

We know that \( \log_{4}(4) = 1 \) because any number to the power of 1 is itself (\(4^1 = 4\)). Therefore, \( \log_{4}(16) = 2 \cdot 1 = 2 \).

Step 4: Substitute Back to Find the Solution

Substitute \( \log_{4}(16) = 2\) back into the expression from Step 1: \( 100 \cdot \log_{4}(16) = 100 \times 2 \).

Step 5: Final Calculation

Calculate \(100 \times 2 = 200\). Thus, \( \log_{4}(16^{100}) = 200 \).

Key Concepts

Exponent Rule for Logarithms

The exponent rule for logarithms is a handy tool that significantly simplifies complex logarithmic expressions. This rule states: \( \log_{b}(a^n) = n \cdot \log_{b}(a) \).
This means when you have a power within a logarithm, you can "bring down" the exponent as a coefficient in front of the logarithm. For example, in the expression \( \log_{4}(16^{100}) \), the exponent 100 can be extracted from the power, allowing us to rewrite it as \( 100 \cdot \log_{4}(16) \).
This transformation makes it easier to compute the logarithm by simplifying the terms involved.To remember this rule, think of it as "power out, product in": the power of the term inside the logarithm is pulled out as a product with the logarithm of the base term. If you keep this in mind, utilizing the exponent rule will become second nature.

Base Conversion

Base conversion in logarithms often simplifies the process of evaluation by reducing expressions into more manageable terms.
To convert the base of the logarithm can be especially useful when the expression inside the log is a power of the base. In our example with \( \log_{4}(16) \), noticing that 16 is actually \( 4^2 \) means we can now express 16 as a power of the base (which is 4 in this case).
Therefore, \( 16 = 4^2 \), enabling our logarithmic expression \( \log_{4}(16) \) to transform into \( \log_{4}(4^2) \).Understanding this step involves recognizing relationships between numbers and knowing basic exponent facts:

• Identify if the number inside the log is a whole power of the base.
• Rewrite the number as a power of the base for easier simplification.

The aim here is to make log expressions resemble something familiar, like \( \log_{b}(b) \), which drastically simplifies calculations.

Simplification of Logarithmic Expressions

Simplifying logarithmic expressions involves breaking them down into basic components that are easily calculable.
After converting the base, the next steps typically involve direct numerical evaluations to simplify further.In continuing from our example, after expressing it as \( \log_{4}(4^2) \), we can apply the simpler expression \( 2 \cdot \log_{4}(4) \).
Knowing that \( \log_{4}(4) = 1 \) since any value logged to its own base equals 1, simplifies \( \log_{4}(4^2) \) to \( 2 \cdot 1 = 2 \).So, simplifying is about leveraging identities like \( \log_{b}(b) = 1 \) to eliminate terms:

• Use known logarithmic identities for direct calculations.
• Directly evaluate simple terms to transform entire expressions quickly.

These simplifications help navigate through complex expressions to find solutions efficiently, turning seemingly tricky problems into straightforward calculations.