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Chapter 13: Problem 49

Solve each exponential equation. $$9^{r}=\frac{1}{27}$$

Short Answer

Expert verified
The solution of the given exponential equation is \(r = -\frac{3}{2}\).

Step by step solution

01

Rewrite the equation using the same base

Given the equation \(9^r = \frac{1}{27}\), we can rewrite both sides of the equation using the base 3 as follows: - 9 can be written as \(3^2\) - 27 can be written as \(3^3\) Thus, the equation becomes \((3^2)^r = \frac{1}{(3^3)}\).
02

Simplify the equation using the exponent rule

Use the power of a power rule to simplify the expression. The rule states that \((a^m)^n = a^{m \times n}\). Applying this rule to the equation gives us: \(3^{2r} = \frac{1}{3^3}\) Now, write the denominator of the right side as an exponent to get: \(3^{2r} = 3^{-3}\)
03

Solve for r using exponent property

Now the equation looks like this: \(3^{2r} = 3^{-3}\). Since both sides of the equation have a base of 3, we can set the exponents equal to each other: \(2r = -3\) Now, solve for r: \(r = -\frac{3}{2}\) So the solution of the given exponential equation is \(r = -\frac{3}{2}\).

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Key Concepts

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

Algebra in Exponential Equations
Algebra is a foundational tool for solving equations, including those with exponential terms. When dealing with an exponential equation like \(9^r = \frac{1}{27}\), our goal is to isolate the variable, often by rewriting the bases to be the same. This process simplifies the equation, making it easier to solve. In algebra, it鈥檚 crucial to understand how operations affect both sides of an equation. For example, changing the bases on both sides allows you to focus purely on the exponents. By performing operations step-by-step and following algebraic principles, complex problems can often become much more manageable.
Understanding Exponent Rules
Exponent rules simplify expressions and solve equations efficiently. The power of a power rule, \((a^m)^n = a^{m \times n}\), is particularly useful. In our example, \((3^2)^r\), is simplified to \(3^{2 \times r}\), which further simplifies to \(3^{2r}\). Other important rules include:
  • The product of powers rule: \(a^m \times a^n = a^{m+n}\)
  • The quotient of powers rule: \(\frac{a^m}{a^n} = a^{m-n}\)
  • The zero exponent rule: \(a^0 = 1\)
By mastering these rules, solving exponential equations becomes much simpler.
Solving Exponential Equations
Solving exponential equations often involves making the equation simpler by using the same base on both sides. After rewriting the initial equation, \((9^r = \frac{1}{27})\) as \((3^{2r} = 3^{-3})\), the problem becomes setting exponents equal because the bases match. This gives \(2r = -3\).
The next step is solving for the variable \(r\). Dividing both sides by 2, we isolate \(r\) to find \(r = -\frac{3}{2}\). By following the outlined steps, you can approach many exponential equations with confidence.
Base Conversion Techniques
Base conversion is key to solving equations with different bases, like turning \(9\) and \(27\) into \(3^2\) and \(3^3\). This conversion is made possible by recognizing that both numbers can be expressed in terms of smaller common bases, in this case, 3.
To do this:
  • Identify the common base.
  • Rewrite each number as an exponent of the common base.
By using base conversion, equations become simpler to manage, making it easier to manipulate and solve.

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Most popular questions from this chapter

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log \sqrt{5}$$

The hydronium ion concentrations, \(\left[\mathrm{H}^{+}\right],\) are given for some common substances. Find the \(\mathrm{pH}\) of each substance (to the tenths place), and determine whether each substance is acidic or basic. $$\text { Egg white: }\left[\mathrm{H}^{+}\right]=2 \times 10^{-8}$$

Solve equation. \(\log _{3} y+\log _{3}(y-8)=2\)

Determine whether each statement is true or false. If it is false, rewrite the statement so that it is true. If \(f^{-1}\) is the inverse of \(f\), then \(\left(f^{-1} \circ f\right)(x)=x\) and \(\left(f \circ f^{-1}\right)(x)=x\)

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log 81$$
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