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Chapter 13: Problem 47

Solve each exponential equation. $$2^{a}=\frac{1}{8}$$

Short Answer

Expert verified
The given exponential equation is \(2^a = \frac{1}{8}\). To solve for 'a', we first rewrite the fraction with a denominator having the same base as the left side, resulting in \(2^a = \frac{1}{2^3}\). We then rewrite the right side as a negative exponent: \(2^a = 2^{-3}\). Since both sides have the same base, their exponents must be equal. Therefore, the solution is \(a = -3\).

Step by step solution

01

Write the fraction with a denominator having the same base as the left side of the equation

To do this, we need to express \(8\) as a power of \(2\), which can be written as \(2^3\). So the equation becomes: $$2^{a}=\frac{1}{2^3}$$
02

Rewrite the right side of the equation as a negative exponent

As per the rules of exponents, \(\frac{1}{x^n} = x^{-n}\). Therefore, the equation becomes: $$2^{a}=2^{-3}$$
03

Solve for the exponent 'a'

Since both sides of the equation have the same base (2), their exponents must be equal. Thus, we can set their exponents equal to each other to find the value of 'a': $$a = -3$$ So the solution to the given exponential equation is \(a = -3\).

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Key Concepts

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

Negative Exponents
Negative exponents might seem confusing at first, but they're actually quite straightforward. Imagine you have the term \(x^{-n}\). The negative exponent indicates that you take the reciprocal of the base. So, \(x^{-n} = \frac{1}{x^n}\). This principle helps in understanding expressions and simplifying equations.

Let's consider why we use negative exponents:
  • They allow us to express division problems as multiplication.
  • They simplify working with fractions in exponential form.
  • They make complex equations more manageable by using consistent rules for multiplication and division.
In the exercise, transforming \(\frac{1}{8}\) into \(2^{-3}\) using negative exponents made it easier to compare both sides of the equation directly.
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true. When you encounter an exponential equation like \(2^a = \frac{1}{8}\), you need techniques to isolate the variable鈥攊n this case, solving for \(a\).

Key steps in solving exponential equations include:
  • Identifying the base of the exponent and ensuring it's consistent on both sides of the equation.
  • Using properties of exponents, such as negative exponents, to rewrite expressions.
  • Equalizing the exponents, since if bases are the same, exponents must be equal to maintain equality.
For the given problem, once both sides had a consistent base, solving the equation came down to equating the exponents, resulting in \(a = -3\). This process demystifies the solution, making it accessible and straightforward.
Laws of Exponents
Laws of exponents are rules that simplify handling exponential expressions. These include the product rule, power rule, and quotient rule, among others. Understanding these rules lets you manipulate exponential terms effectively.

Here's a quick rundown of some essential laws:
  • Product of Powers: \(x^m \cdot x^n = x^{m+n}\).
  • Quotient of Powers: \(\frac{x^m}{x^n} = x^{m-n}\).
  • Power of a Power: \((x^m)^n = x^{m \cdot n}\).
  • Negative Exponent: \(x^{-n} = \frac{1}{x^n}\).
Applying these laws reduces complex algebraic expressions to simpler forms. In this particular exercise, rewriting \(\frac{1}{8}\) as \(2^{-3}\) was a demonstration of using these laws to solve for the variable. This ability to maneuver within exponential expressions is essential for more advanced math courses too.
Equivalent Expressions
In mathematics, different expressions can reflect the same value but look different at first glance. These are known as equivalent expressions. Rewriting an expression into a form with shared base and exponents can reveal its equivalence, facilitating easier solving of equations.

Key points involving equivalent expressions include:
  • Changing the form of numbers while keeping their value the same, like expressing 8 as \(2^3\).
  • Applying the laws of exponents to simplify or transform expressions. This includes using negative exponents or fractional exponents as necessary.
  • Creating a path to solve equations by equating similar terms.
In our example, transforming \(\frac{1}{8}\) into \(2^{-3}\) is a textbook case of finding equivalent expressions that allow easier solution of the equation \(2^a = 2^{-3}\). Such transformations make solving equations more accessible, highlighting the power of manipulating equivalent expressions in math.

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Most popular questions from this chapter

Use the formula \(A=P\left(1+\frac{r}{n}\right)^{n t}\) to solve each problem. Find the amount Christopher owes at the end of 5 yr if he borrows \(\$ 4000\) at a rate of \(6.5 \%\) compounded quarterly.

Write as a single logarithm. Assume the variables are defined so that the variable expressions are positive and so that the bases are positive real numbers not equal to \(1 .\) $$4 \log _{3} f+\log _{3} g$$

Show that the inverse of \(y=\ln x\) is \(y=e^{x}\).

Find the inverse of each one-to-one function. Then, graph the function and its inverse on the same axes. $$f(x)=x^{3}$$

Find the inverse of each one-to-one function. $$g(x)=-4 x+8$$
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