Chapter 5: Q 100. (page 309)

Find the value of $\log_{2} 2 \cdot \log_{2} 4 \cdot \dots \cdot \log_{2} 2^{n}$ .

Short Answer

Expert verified

The result is $n!$ .

Step by step solution

Step 1. Given information.

Consider the given expression.

$\log_{2} 2 \cdot \log_{2} 4 \cdot \dots \cdot \log_{2} 2^{n}$

Step 2. Calculation.

Apply the base change formula to simplify the expression.

$\begin{array}{rcl} \log_{2} 2 \cdot \log_{2} 4 \cdot \dots \cdot \log_{2} 2^{n} & = & \frac{\log 2}{\log 2} \cdot \frac{\log 4}{\log 2} \cdot \dots \cdot \frac{\log 2^{n}}{\log 2} \\ = & \frac{\log 2^{1}}{\log 2} \cdot \frac{\log 2^{2}}{\log 2} \cdot \dots \cdot \frac{n \log 2}{\log 2} \\ = & 1 \cdot 2 \cdot 3 \cdot \dots \cdot n \\ = & n! \end{array}$

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Find the value of $\log_{2} 2 \cdot \log_{2} 4 \cdot \dots \cdot \log_{2} 2^{n}$ .

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Calculation.

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Calculation.

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Find the value of $\log_{2} 2 \cdot \log_{2} 4 \cdot \dots \cdot \log_{2} 2^{n}$ .