Problem 3 - Precalculus: Functions and Graphs

Question

Change to exponential form. (a) \(\log _{2} 32=5\) (b) \(\log _{3} \frac{1}{243}=-5\) (c) \(\log _{t} r=p\) (d) \(\log _{3}(x+2)=5\) (e) \(\log _{2} m=3 x+4\) (f) \(\log _{4} 512=\frac{3}{2}\)

Step-by-Step Solution

Answer

Exponential forms are: (a) \(2^5 = 32\), (b) \(3^{-5} = \frac{1}{243}\), (c) \(t^p = r\), (d) \(3^5 = x+2\), (e) \(2^{3x+4} = m\), (f) \(4^{\frac{3}{2}} = 512\).

Step 1: Understand Logarithmic Form

The logarithmic form \( \log_b a = c \) can be understood as the power \( c \) to which the base \( b \) must be raised to get \( a \). We need to convert these into exponential form \( b^c = a \).

Step 2: Convert (a) to Exponential Form

Given \( \log _{2} 32=5 \), the base \( 2 \), the argument \( 32 \), and the exponent \( 5 \). The exponential form is \( 2^5 = 32 \).

Step 3: Convert (b) to Exponential Form

Given \( \log _{3} \frac{1}{243}=-5 \), the base \( 3 \), the argument \( \frac{1}{243} \), and the exponent \( -5 \). The exponential form is \( 3^{-5} = \frac{1}{243} \).

Step 4: Convert (c) to Exponential Form

Given \( \log _{t} r=p \), the base \( t \), the argument \( r \), and the exponent \( p \). The exponential form is \( t^p = r \).

Step 5: Convert (d) to Exponential Form

Given \( \log _{3}(x+2)=5 \), the base \( 3 \), the argument \( x+2 \), and the exponent \( 5 \). The exponential form is \( 3^5 = x + 2 \).

Step 6: Convert (e) to Exponential Form

Given \( \log _{2} m=3x+4 \), the base \( 2 \), the argument \( m \), and the exponent \( 3x+4 \). The exponential form is \( 2^{3x + 4} = m \).

Step 7: Convert (f) to Exponential Form

Given \( \log _{4} 512=\frac{3}{2} \), the base \( 4 \), the argument \( 512 \), and the exponent \( \frac{3}{2} \). The exponential form is \( 4^{\frac{3}{2}} = 512 \).

Key Concepts

Logarithmic Form

A logarithm is another way of expressing an exponentiation relationship between numbers. In a logarithmic expression, such as \( \log_b a = c \), we think of it in terms of an inverse operation to exponentiation. It tells us that the base \( b \) raised to the power of \( c \) equals \( a \). This form is useful for solving equations where the exponent is unknown, and it can easily be converted into exponential form—which is often more intuitive to understand.

• Logarithmic form reveals the exponent or power needed for a base to achieve a particular number.
• It's especially handy for solving problems involving exponential growth, decay, or transformations.

Base and Exponent

In both exponential and logarithmic forms, the base and exponent play vital roles. The base \( b \) in an expression \( b^c = a \) is the number that is multiplied. It is raised to the power of the exponent \( c \). In logarithmic form \( \log_b a = c \), the base \( b \) is the same number used to determine what power (exponent) is needed to reach the argument \( a \).Understanding these components helps in converting between logarithmic and exponential forms:

• Base: The number being repeatedly multiplied (in exponential form) or the number you're asking about (in logarithmic form).
• Exponent: Indicates the number of times the base is multiplied by itself. In logarithmic terms, it's the result you're solving for.

These elements help us manage calculations involving large numbers or repeated multiplications by turning them into manageable addition operations through logarithms.

Argument

The argument in logarithmic form, \( \log_b a = c \), refers to the value \( a \) that results from raising the base \( b \) to the power of \( c \). It is a crucial part of the expression because it indicates the actual result or product of the base powered by the exponent. In converting from logarithmic to exponential form, the argument becomes the number that the operations are centered upon.

• The argument represents the outcome of exponential growth or decay.
• It allows us to reverse the process of exponentiation through logarithms.

By understanding the argument, we can determine the effectiveness of using logarithms to simplify or solve exponential equations, making it easier to handle problems where the result is known, and the base or exponent is unknown.