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Chapter 3: Problem 4

Solve for \(x\) algebraically. $$9^{x}=\frac{1}{243}$$

Short Answer

Expert verified
The solution to the equation \(9^{x} = \frac{1}{243}\) is \(x = -3\)

Step by step solution

01

Write the equation

First, rewrite the equation \(9^{x} = \frac{1}{243}\)
02

Change the base on the right side

Next, recognize that \(243=9^3\), so that the base can be expressed as 9. So, the equation becomes \(9^{x} = 9^{-3}\), thanks to the property that says, \(\frac{1}{a^n} = a^{-n}\)
03

Equate Exponents of Same Base

According to the rule of exponents, when we have equal bases we can set the exponents to be equal. Therefore, \(x = -3\)

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Key Concepts

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

Solving for x
When faced with the task of solving for \(x\) in an exponential equation, our goal is to determine the value that makes the equation true. The equation given is \(9^{x} = \frac{1}{243}\). To solve for \(x\), it is essential to simplify and manipulate the equation using algebraic methods and properties of exponents. This often involves expressing both sides of the equation with a common base. Then, you can use properties of exponents to equate the powers, leading directly to the solution. The critical skill here is to identify the base and adjust it to simplify finding \(x\).
Properties of Exponents
Understanding the properties of exponents is key to solving exponential equations like \(9^{x} = \frac{1}{243}\). Here are some essential properties:
  • Product of Powers: \(a^m \times a^n = a^{m+n}\)
  • Power of a Power: \((a^m)^n = a^{m \cdot n}\)
  • Negative Exponents: \(a^{-n} = \frac{1}{a^n}\)

Changing \(\frac{1}{243}\) to \(9^{-3}\) involved recognizing \(243\) as \(9^3\). From here, we applied the property of negative exponents to express \(1/9^3\) as \(9^{-3}\). These properties serve as the tools to transform complex equations into simpler ones that can be more easily solved.
Algebraic Methods
Algebraic methods in solving exponential equations focus on transforming the equation to allow \(x\) to be isolated and solved. In equation \(9^{x} = \frac{1}{243}\), the strategy is to set the bases equal. Once achieved, the exponents can be equated just like regular numbers.
Here's how it progresses:
  • Transform the equation into a common base, \(9\) in this case\(,\) using exponent properties.
  • Because both sides share the same base, set the exponents equal: \(x = -3\).
This manipulation ensures we directly solve the equation algebraically, using justifications based on known properties of equations. It highlights how each algebraic step is justified via exponent rules and the setting of equal bases.

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Most popular questions from this chapter

The Rhind papyrus, named after A. Henry Rhind, contains most of what we know today of ancient Egyptian mathematics. A chemical analysis of a sample from the papyrus has shown that it contains approximately \(75 \%\) of its original carbon-14. What is the age of the Rhind papyrus? Use 5730 years as the halflife of carbon- 14

Solve \(0.5=e^{14 k}\) for \(k .\) Round to the nearest ten-thousandth. [3.5]

Determine the domain of the given function. Write the domain using interval notation. $$f(x)=\sqrt{1-e^{x}}$$

Make use of the factorial function, which is defined as follows. For whole numbers \(n\), the number \(n !\) (which is read "n factorial") is given by $$n !=\left\\{\begin{array}{ll} n(n-1)(n-2) \cdots 1, & \text { if } n \geq 1 \\\1, & \text { if } n=0 \end{array}\right.$$Thus, \(0 !=1\) and \(4 !=4 \cdot 3 \cdot 2 \cdot 1=24\) STIRLING'S FORMULA Stirling's Formula (after James Stir\(\operatorname{lin} g, 1692-1770)\) 1$$n !=\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}$$ is often used to approximate very large factorials. Use Stirling's Formula to approximate \(10 !\), and then compute the ratio of Stirling's approximation of \(10 !\) divided by the actual value of \(10 !\), which is 3,628,800

Starting on the left side of a standard 88 -key piano, the frequency, in vibrations per second, of the \(n\) th note is given by \(f(n)=(27.5) 2^{(n-1) / 12}\) a. Using this formula, determine the frequency, to the nearest hundredth of a vibration per second, of middle C, key number 40 on an 88 -key piano. b. Is the difference in frequency between middle C (key number 40 ) and \(D\) (key number 42 ) the same as the difference in frequency between \(\mathrm{D}\) (key number 42 ) and \(\mathrm{E}\) (key number 44 )? Explain.
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