Problem 72 - Algebra 2 and Trigonometry

Question

If \(\mathrm{h}(x)=\log _{32} x,\) find \(\mathrm{h}(8)\)

Step-by-Step Solution

Answer

\( \mathrm{h}(8) = \frac{3}{5} \)

Step 1: Understand the Function

We have the function \( \mathrm{h}(x) = \log_{32} x \). This is a logarithmic function with base 32.

Step 2: Substitute the Value

We want to find \( \mathrm{h}(8) \), so we substitute 8 for \( x \) in the function: \( \mathrm{h}(8) = \log_{32} 8 \).

Step 3: Convert to a Base 2 Logarithm

To simplify \( \log_{32} 8 \), we convert it to a base-2 logarithm. Since \( 32 = 2^5 \) and \( 8 = 2^3 \), we can use the change of base formula: \[ \log_{32} 8 = \frac{\log_{2} 8}{\log_{2} 32} \].

Step 4: Simplify Logarithms

Calculate the base-2 logarithms: \( \log_{2} 8 = 3 \) and \( \log_{2} 32 = 5 \). Substitute these values into the previous expression: \[ \log_{32} 8 = \frac{3}{5} \].

Key Concepts

Change of Base Formula

The change of base formula is a nifty tool in mathematics that helps us calculate logarithms more easily when dealing with bases other than 10 or e. It allows a logarithm to be expressed in terms of logarithms with a different base, often simplifying calculations. To use the change of base formula, you take a logarithm with an uncommon base and express it as a fraction of two logarithms with a common base, usually 10 or e, but it works with any base you choose.

Here's the formula:

• For any logarithm \( \log_b a \), it can be rewritten as: \( \log_b a = \frac{\log_k a}{\log_k b} \)
• This means \( a \) over some base \( b \) can be broken down to a division of two logarithms with a new base \( k \).

The choice of \( k \) could be any number, but using convenient bases like 10 or 2 makes calculations smoother. In our example, the calculation of \( \log_{32} 8 \) became simpler by converting it to base 2 logarithms. This approach helps in both simplifying log expressions and solving complex logarithmic equations.

Base Conversion

Base conversion is an essential part of understanding and working with logarithms. A number's base determines how it's expressed in a different numbering system. For example, binary (base 2), decimal (base 10), and hexadecimal (base 16) are common numbering systems. In logarithms, different bases can make calculations cumbersome, which is why base conversion becomes handy.

The connection to logarithms is that when we convert a logarithm from one base to another, we're essentially translating the problem to a format that may be easier to handle. In the given exercise, the base conversion comes into play as we see that:

• The base 32 and 8 can be expressed in terms of base 2, specifically as powers: \( 32 = 2^5 \) and \( 8 = 2^3 \).
• This translation allows us to apply the change of base formula successfully, making computations much simpler.

When dealing with bases in logarithms, being comfortable with changing and converting among them is a powerful technique for simplifying complex equations and understanding them more clearly.

Logarithmic Equations

Logarithmic equations involve variables in the position of the logarithmic argument or sometimes the base. Solving these equations requires a solid understanding of the properties of logarithms and often includes manipulating the equations to a form that allows for direct solution or simplification.

One common scenario is when you need to solve for the argument or convert bases to simplify the equation. Here’s a straightforward approach:

• Identify the logarithmic form. Isolate the logarithm if possible, to make comparison or conversion easy.
• Apply the change of base or logarithmic properties to make the expression straightforward, possibly transforming it into a linear form.
• Once simplified, solve the resulting equation for the unknown.

In our specific logarithmic equation problem, \( h(x) = \log_{32} x \), to find \( h(8) \), we applied the change of base to derive the simpler equation:\( h(8) = \frac{3}{5} \). Recognizing the conditions where logarithmic properties simplify calculations can greatly enhance problem-solving efficiency.