Problem 41 - Precalculus

Question

Evaluate the logarithms exactly (if possible). $$\log _{1 / 4} 4096$$

Step-by-Step Solution

Answer

The value of \( \log_{1/4} 4096 \) is -6.

Step 1: Understand the Logarithm Expression

The given expression is \( \log_{1/4} 4096 \). This represents the exponent to which the base \( \frac{1}{4} \) must be raised to obtain 4096. Our goal is to find this exponent.

Step 2: Rewrite the Base as a Power of 2

Recognize that \( \frac{1}{4} \) can be rewritten as a power of 2. Since \( 4 = 2^2 \), it follows that \( \frac{1}{4} = 2^{-2} \). Therefore, the base \( \frac{1}{4} \) is equivalent to \( 2^{-2} \).

Step 3: Rewrite the Argument as a Power of 2

Recognize that 4096 can also be expressed as a power of 2. Since \( 4096 = 2^{12} \), we re-write the argument as \( 2^{12} \).

Step 4: Set Up the Equation

Now substitute the rewritten base and argument into the log expression: \( \log_{1/4} 4096 \) becomes \( \log_{2^{-2}} 2^{12} \). This is equivalent to finding \( x \) such that \( (2^{-2})^x = 2^{12} \).

Step 5: Use Properties of Exponents

Using properties of exponents, we know that \( (a^b)^c = a^{bc} \). Therefore, \( (2^{-2})^x = 2^{12} \) implies \( 2^{-2x} = 2^{12} \).

Step 6: Set Exponents Equal to Each Other

Set the exponents equal because the bases are the same: \( -2x = 12 \).

Step 7: Solve for x

Solve the equation \( -2x = 12 \) for \( x \). Divide both sides by -2: \( x = \frac{12}{-2} = -6 \). Thus, \( x = -6 \).

Key Concepts

Understanding Exponents

Exponents are mathematical expressions that represent repeated multiplication of a number by itself. For example, in the expression \( a^b \), \( a \) is the base and \( b \) is the exponent.
The expression signifies that \( a \) is multiplied by itself \( b \) times.

• If \( b = 2 \), then \( a^b \) is called "a squared."
• If \( b = 3 \), then it's referred to as "a cubed."
• If \( b "> 3 \), we simply say "a to the power of b."

When dealing with negative exponents, the expression \( a^{-b} \) is equivalent to \( \frac{1}{a^b} \). This helps in converting expressions with negative exponents into fractions, which is crucial in solving logarithmic problems like the one given.

Base Conversion in Logarithms

Base conversion involves changing one base into another to simplify computation. Using the exercise, we converted the base \( \frac{1}{4} \) into a power of 2, which is more manageable. Since \( 4 = 2^2 \), the inverse \( \frac{1}{4} \) can be rewritten as \( 2^{-2} \).
This reexpression allows us to view logarithms in terms of a familiar base, contributing to easier problem-solving.

• Conversion simplifies complex calculations, making it easier to handle powers of familiar bases like 2, 10, or e.
• It aids in identifying the exponent relationships more directly, which is essential in solving logarithmic equations.

Recognizing opportunities for base conversion can significantly ease the calculation process.

Properties of Exponents

The properties of exponents are rules that govern the operations with exponential expressions.
These are crucial when manipulating powers in logarithmic expressions.

• Product of Powers: \( a^m \times a^n = a^{m+n} \).
• Power of a Power: \( (a^m)^n = a^{m \times n} \).
• Quotient of Powers: \( \frac{a^m}{a^n} = a^{m-n} \).

In the problem, we applied **Power of a Power**:

When you have \( (a^b)^c \), it translates into \( a^{bc} \).
This enabled us to equate \( 2^{-2x} \) with \( 2^{12} \), leading directly to solving for \( x \) by making the exponents equal. Understanding these properties helps in simplifying expressions and finding exact solutions.

Solving Exponential Equations

Exponential equations involve expressions where the variable is located in the exponent. To solve such equations effectively, understanding both the base and exponents is key.

• To simplify, align the equation so that both sides have the same base.
• Once the bases are equal, set the exponents equal to solve for the variable.
• In this exercise, the equation \( (2^{-2})^x = 2^{12} \) transitions to solving \(-2x = 12\).

By converting \( -2x = 12 \), we divide both sides by -2 to find \( x \).
This logical breakdown assists in clearly understanding the role of exponents and helps in accurately determining the unknown variable. Mastering exponential equations opens up possibilities in handling both simple and complex logarithmic expressions.

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Other exercises in this chapter

Problem 41

Write each expression as a single logarithm. $$\frac{1}{2} \log _{b} x+\frac{2}{3} \log _{b} y$$

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Problem 41

Graph the exponential function using transformations. State the \(y\) -intercept, two additional points, the domain, the range, and the horizontal asymptote. $$

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Problem 42

Solve the logarithmic equations exactly. $$\log _{2}(3 x-1)=3$$

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Problem 42

Write each expression as a single logarithm. $$\frac{1}{2} \log _{b} x-\frac{2}{3} \log _{b} y$$

View solution