Problem 33 - College Algebra

Question

For the following exercises, use the definition of a logarithm to solve the equation. $$ 4+\log _{2}(9 k)=2 $$

Step-by-Step Solution

Answer

The solution to the equation is \( k = \frac{1}{36} \).

Step 1: Isolate the Logarithmic Part

Start by isolating the logarithm on one side of the equation. Given the equation \( 4 + \log_{2}(9k) = 2 \), subtract 4 from both sides to obtain: \[ \log_{2}(9k) = 2 - 4 \]This simplifies to:\[ \log_{2}(9k) = -2 \]

Step 2: Rewrite the Logarithmic Equation to Exponential Form

Use the definition of a logarithm to rewrite the equation \( \log_{b}(a) = c \) as \( a = b^{c} \). Here, we have:\[ 9k = 2^{-2} \]

Step 3: Solve the Exponential Equation

Simplify the right side of the equation. Calculate the value of \( 2^{-2} \):\[ 2^{-2} = \frac{1}{2^{2}} = \frac{1}{4} \]So we have:\[ 9k = \frac{1}{4} \]

Step 4: Solve for k

Divide both sides of the equation \( 9k = \frac{1}{4} \) by 9 to solve for \( k \):\[ k = \frac{1}{4 \times 9} = \frac{1}{36} \]

Key Concepts

Exponential Form

When solving logarithmic equations, understanding the exponential form can be incredibly useful. In mathematics, the concept of exponential form provides a way to express a number as a power of another number. This is typically shown as \( a = b^{c} \). The 'base' \( b \) is raised to the 'exponent' \( c \) to get the result \( a \). This notion is critical as it gives a simpler and more intuitive way of understanding equations that involve exponents.

In the original problem, we had to convert the logarithmic equation \( \log_{2}(9k) = -2 \) into its exponential form to simplify solving it. Using the rule, a logarithm \( \log_{b}(a) = c \) implies that \( a = b^{c} \), we express \( 9k \) as \( 2^{-2} \).

Converting from a logarithmic to exponential form is often the key step in making logarithmic equations easier to solve, as it transforms a complex log operation into a simpler multiplication.

Solving Equations

Solving equations, especially when involving logs and exponentials, is about isolating variables to find their values. The objective is to manipulate the equation so that the variable is alone on one side.

Here's a brief approach on how to tackle such questions:

• Isolate the logarithmic expression. In our problem, we did this by subtracting 4 from both sides, which simplified it to \( \log_{2}(9k) = -2 \).
• Convert the logarithmic equation to exponential form, which we achieved by writing \( 9k \) as \( 2^{-2} \).
• Expand the exponential term if necessary. We simplified \( 2^{-2} \) to \( \frac{1}{4} \) because it helps make further calculations clearer.
• Solve for the variable. In this case, we divided both sides by 9 to find \( k \).

These steps create a systematic path to unravel even the most complex looking equations.

Logarithmic Definitions

Logarithms may seem tricky at first, but understanding their definitions can demystify the concept and make problems much easier to solve. The basic idea of a logarithm \( \log_{b}(a) = c \) is that it asks the question: "To what power should we raise \( b \) to get \( a \)?">

For example, in our problem, \( \log_{2}(9k) = -2 \) translates to asking "To what power should 2 be raised to get \( 9k \)?" The answer is \( -2 \), which means \( 9k = 2^{-2} \).

Here are some important points to remember about logarithms:

• They are the inverse operations of exponentials. If \( a = b^{c} \), then \( c = \log_{b}(a) \).
• The base \( b \) must be a positive number, not equal to 1.
• Logarithms can be used to solve for unknown quantities in exponential equations.

Understanding these basic principles of logarithms is essential for solving problems that involve complex exponential and logarithmic expressions.