Question

Find the value of \(\log _{2} 4 \cdot \log _{4} 6 \cdot \log _{6} 8\).

Short answer

The value is 3.

Step-by-step solution

Simplify \(\texttt{log}_{2} 4\)

Recognize that 4 is a power of 2. Specifically, \(4 = 2^2\). Therefore, \(\texttt{log}_{2} 4 = \texttt{log}_{2} (2^2)\). Using the property of logarithms \( \texttt{log}_{b} (a^c) = c \texttt{log}_{b} a \), we get \( \texttt{log}_{2} (2^2) = 2 \texttt{log}_{2} 2 = 2 \).

Simplify \(\texttt{log}_{4} 6\)

Using the change of base formula \( \texttt{log}_{b} a = \frac{\texttt{log}_{c} a}{\texttt{log}_{c} b} \), we can rewrite \( \texttt{log}_{4} 6 = \frac{\texttt{log}_{2} 6}{\texttt{log}_{2} 4} \). From Step 1, we know \( \texttt{log}_{2} 4 = 2 \), so \( \texttt{log}_{4} 6 = \frac{\texttt{log}_{2} 6}{2} \).

Simplify \(\texttt{log}_{6} 8\)

Using the change of base formula again, we rewrite \( \texttt{log}_{6} 8 = \frac{\texttt{log}_{2} 8}{\texttt{log}_{2} 6} \). Notice that \( 8 = 2^3 \), so \(\texttt{log}_{2} 8 = \texttt{log}_{2} (2^3) = 3 \). Hence, \( \texttt{log}_{6} 8 = \frac{3}{\texttt{log}_{2} 6} \).

Combine all terms

We now have:1. \(\texttt{log}_{2} 4 = 2\)2. \(\texttt{log}_{4} 6 = \frac{\texttt{log}_{2} 6}{2}\)3. \(\texttt{log}_{6} 8 = \frac{3}{\texttt{log}_{2} 6}\)Combine these results:\( \texttt{log}_{2} 4 \cdot \ \texttt{log}_{4} 6 \cdot \ \texttt{log}_{6} 8 = 2 \cdot \ \frac{\texttt{log}_{2} 6}{2} \cdot \ \frac{3}{\texttt{log}_{2} 6} \).

Simplify the expression

Notice that \( \texttt{log}_{2} 6 \) in the numerator and denominator cancels out, and \( 2 \) in the numerator and denominator also cancels out:\( 2 \cdot \ \frac{\texttt{log}_{2} 6}{2} \cdot \ \frac{3}{\texttt{log}_{2} 6} = 2 \cdot \ 1 \cdot \ 3 = 3 \).Therefore, the value of the expression is 3.

Key concepts

Change of Base Formula

The change of base formula is a useful tool in logarithms that allows you to rewrite a logarithm in terms of logs with a different base.
The formula is: \[ \text{log}_b a = \frac{\text{log}_c a}{\text{log}_c b} \]
This means you can express \( \text{log}_{4} 6 \) as \( \frac{\text{log}_{2} 6}{\text{log}_{2} 4} \).
We often use the change of base formula when we do not have a calculator that directly computes logarithms of the given base.
Instead, we convert to a base, like base 10 or base 2, which is typically supported by calculators.

Logarithm Simplification

Simplifying logarithms often involves recognizing powers and using the properties of logarithms.
For instance, \( \text{log}_2 4 \) simplifies because 4 is \( 2^2 \). Here, you can apply the property: \[ \text{log}_b (a^c) = c \text{log}_b a \]
This tells us that \( \text{log}_2 (2^2) = 2 \text{log}_2 2 \).
Since \( \text{log}_2 2 \) equals 1, we simplify to 2.
Recognizing these patterns is key to making logarithmic expressions more manageable.

Logarithmic Identities

Understanding logarithmic identities can simplify complex logarithmic expressions.
Some key identities include:

• \( \text{log}_b (xy) = \text{log}_b x + \text{log}_b y \)
• \( \text{log}_b \frac{x}{y} = \text{log}_b x - \text{log}_b y \)
• \( \text{log}_b (a^c) = c \text{log}_b a \)

For example, knowing \( \text{log}_b a = \frac{\text{log}_c a}{\text{log}_c b} \) allowed us to rewrite \( \text{log}_4 6 \) in terms of logs base 2.

Logarithm Multiplication

Multiplying logarithms involves using properties and identities to simplify the expression.
In the example \( \text{log}_{2} 4 \times \text{log}_{4} 6 \times \text{log}_{6} 8 \), handle each log term individually:

• Simplify \( \text{log}_{2} 4 \)
• Convert \( \text{log}_{4} 6 \) using the change of base formula
• Simplify \( \text{log}_{6} 8 \) using the change of base formula and recognize powers

Finally, combine and simplify: \( 2 \times \frac{\text{log}_{2} 6}{2} \times \frac{3}{\text{log}_{2} 6} \), and the expression simplifies to 3 through cancellations.