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Chapter 6: Problem 5

In Exercises \(1-33,\) solve the equation analytically. $$ 8^{x}=\frac{1}{128} $$

Short Answer

Expert verified
The solution is \( x = \frac{-7}{3} \).

Step by step solution

01

Express in Terms of Same Base

The given equation is \( 8^x = \frac{1}{128} \). We need to express both sides with the same base of 2. Recognize that \( 8 = 2^3 \) and \( 128 = 2^7 \). Also, since \( \frac{1}{128} \) can be rewritten as \( 2^{-7} \), our equation becomes \( (2^3)^x = 2^{-7} \).
02

Apply the Power of a Power Rule

Using the power of a power rule \((a^m)^n = a^{m \times n}\), rewrite \( (2^3)^x \) as \( 2^{3x} \). Now our equation is \( 2^{3x} = 2^{-7} \).
03

Set the Exponents Equal

Since the bases are equal, we can set the exponents equal to each other, leading to the equation \( 3x = -7 \).
04

Solve for x

Solve the equation \( 3x = -7 \) by dividing both sides by 3 to isolate \( x \). This gives us \( x = \frac{-7}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power of a Power Rule
The power of a power rule is a fundamental principle in exponentiation, which deals with expressions like \((a^m)^n\). The rule states that when you raise a power to another power, you multiply the exponents together, resulting in \(a^{m \times n}\). This concept helps simplify complex expressions and solve exponential equations more efficiently.
  • Consider a base \(a\) raised to a power \(m\). This entire expression \((a^m)\) is then raised to another power \(n\).
  • According to this rule, multiply the exponents: \(m \times n\).
  • The new expression becomes \(a^{m \times n}\).
Applying this to our exercise, we had \((2^3)^x\). By using the power of a power rule, it becomes \(2^{3x}\). This simplifies the equation and allows us to equate the exponents directly when the bases are the same.
Exponentiation
Exponentiation involves raising a number, called the base, to a power, which is an exponent. In its essence, exponentiation is repeated multiplication of the base.
  • For example, \(2^3\) means multiplying 2 by itself 3 times: \(2 \times 2 \times 2 = 8\).
  • The base is the number being multiplied, and the exponent signifies how many times the base is used in the multiplication.
  • Exponentiation is a crucial operation for solving equations, especially when variables are present in the exponents.
In the exercise, both the base and the exponent were manipulated using base conversion and exponent rules. Understanding exponentiation is key to grasping these changes and solving the equation effectively.
Base Conversion Techniques
Base conversion techniques help transform numbers or equations into a common base, allowing for simpler computation and comparison. When numbers are converted to the same base, it is easier to solve equations where the bases must be equal for a direct comparison of exponents.
  • In our exercise, the numbers 8 and 128 were converted to a base of 2 because both are powers of 2.
  • Conversion: \(8\) was expressed as \(2^3\), and \(128\) as \(2^7\).
  • This technique made it possible to transform the given problem into \((2^3)^x = 2^{-7}\), facilitating an easier route to solving the equation.
By mastering base conversion techniques, students can tackle more complex expressions confidently, as it permits direct comparisons and simplifications in exponential equations.

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Most popular questions from this chapter

We list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula \(A(t)=A_{0} e^{k t}\) where \(A_{0}\) is the initial amount of the material and \(k\) is the decay constant. For each isotope: \- Find the decay constant \(k\). Round your answer to four decimal places. \- Find a function which gives the amount of isotope \(A\) which remains after time \(t\). (Keep the units of \(A\) and \(t\) the same as the given data.) \- Determine how long it takes for \(90 \%\) of the material to decay. Round your answer to two decimal places. (HINT: If \(90 \%\) of the material decays, how much is left?) Uranium 235 , used for nuclear power, initial amount \(1 \mathrm{~kg}\) grams, half-life 704 million years.

We introduce three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay attention to the subtle differences. While the decibel scale can be used in many disciplines, \(^{13}\) we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound. \(^{14}\) The Sound Intensity Level \(L\) (measured in decibels) of a sound intensity \(I\) (measured in watts per square meter) is given by $$L(I)=10 \log \left(\frac{I}{10^{-12}}\right)$$ Like the Richter scale, this scale compares \(I\) to baseline: \(10^{-12} \frac{W}{m^{2}}\) is the threshold of human hearing. (a) Compute \(L\left(10^{-6}\right)\). (b) Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity \(I\) is needed to produce this level? (c) Compute \(L(1)\). How does this compare with the threshold of pain which is around 140 decibels?

In Exercises 1 - 15 , expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers. $$ \ln \left(x^{3} y^{2}\right) $$

Evaluate the expression. \(\log _{4}(8)\)

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