/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Solve the logarithmic equation a... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 34

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log x=1$$

Short Answer

Expert verified
The given logarithmic equation is \(\log x = 1\). To solve for x, we first convert it to exponential form: \(10^1 = x\). Therefore, we find that \(x = 10\). To check the solution using a graphing calculator, enter the function \(y = \log x\) and verify that the graph passes through the point (10,1). This confirms our algebraic solution x = 10.

Step by step solution

01

Convert logarithmic equation to exponential form

First, we need to rewrite the given logarithmic equation \(\log x = 1\) into an equivalent exponential form. Since the logarithm has no base written, we assume it has base 10. So, we have: \[10^1 = x\]
02

Solve for x

Now, we solve the equation for x: \[x = 10^1\] \[x = 10\] So the solution to the given logarithmic equation is x = 10.
03

Checking the solution using a graphing calculator

To check our solution using a graphing calculator, we'll follow these steps: 1. Turn on the graphing calculator. 2. Choose a mode for graphing functions (i.e., function or y=). 3. Enter the function \(y = \log x\). 4. Graph the function and locate the point where y = 1. 5. Read the x-coordinate of that point. It should match our algebraic solution. If the graph of the function \(y = \log x\) passes through the point (10,1), then our algebraic solution x = 10 is correct.

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Key Concepts

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

Exponential Form
When solving logarithmic equations, converting them to exponential form often simplifies the process. A logarithmic expression like \(\log_{10} x = y\) can be rewritten using the concept of an exponent. This converts it into \(10^y = x\). For example, solving \(\log x = 1\) means finding a number \(x\) such that \(10^1 = x\). Therefore, the exponential form is \(x = 10\). This algebraic rearrangement helps students see logarithms as exponents, making them easier to visualize and manipulate. Creating a clear connection between logarithms and exponents is crucial for understanding more complex math problems later on.
Graphing Calculator
Checking your solution with a graphing calculator is a useful way to confirm the answer you've found algebraically. Below are the steps you can take using a graphing calculator:
  • Turn on the device and select the graphing mode, often labeled as 'function' or 'y=' mode.
  • Input the logarithmic function, such as \(y = \log x\), into the calculator.
  • Create a graph of this function on the calculator screen.
  • Look for the y-coordinate corresponding to 1, and find the x-coordinate at this point.
  • This x-value should match your earlier algebraic solution, confirming accuracy.
By using visual graphs, you gain a deeper understanding of the equation's behavior and how solutions manifest on a graph, reinforcing your learning through visual representation.
Algebraic Solution
The algebraic method for solving \(\log x = 1\) uses manipulation and understanding of logarithms and exponents. When the exercise presents \(\log x = 1\), it implies using base 10, making the equation \(\log_{10} x = 1\). To isolate \(x\), you then switch the format of the equation to exponential form, resulting in \(10^1 = x\), which straightforwardly gives \(x = 10\).
The algebraic solution benefits from the step-by-step transformation of the equation:
  • Transform \(\log x = 1\) to \(10^1 = x\).
  • Simplify the expression \(10^1\) to find \(x = 10\).
Such direct algebraic manipulation helps cement the understanding that logarithms and exponents are intertwined operations.
Base 10 Logarithm
A base 10 logarithm, often written simply as \(\log\), is a logarithm with 10 as its base. Commonly used in many mathematical and scientific calculations, it helps to solve equations involving powers of 10. For instance, solving \(\log x = 1\) uses the property of base 10 logarithms: \(\log_{10} x = 1\) translates to \(10^1 = x\).
Base 10 is known as the "common logarithm" because of its frequent use in measurement and scientific applications. Understanding this base allows you to seamlessly tackle equations by leveraging its properties:
  • It simplifies operations involving large numbers due to its scale.
  • Recognizes patterns in data that have exponential growth or decay.
  • Facilitates calculations with technology, as seen with the use of calculators.
Mastering base 10 logarithms provides a critical foundation for higher-level math and practical problem-solving capabilities.

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Most popular questions from this chapter

One 8-oz serving each of brewed coffee, Red Bull energy drink, and Mountain Dew soda contains a total of \(197 \mathrm{mg}\) of caffeine. One serving of brewed coffee has \(6 \mathrm{mg}\) more caffeine than two servings of Mountain Dew. One serving of Red Bull contains 37 mg less caffeine than one serving each of brewed coffee and Mountain Dew. (Source: Australian Institute of Sport) Find the amount of caffeine in one serving of each beverage.

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{2}(10+3 x)=5$$

Use a graphing calculator to find the approximate solutions of the equation. $$\log _{5}(x+7)-\log _{5}(2 x-3)=1$$

The top three apple growers in the world - China, the United States, and Turkey - grew a total of about 74 billion lb of apples in a recent year. China produced 44 billion lb more than the combined production of the United States and Turkey. The United States produced twice as many pounds of apples as Turkey. (Source: U.S. Apple Association) Find the number of pounds of apples produced by each country. (THE IMAGES CANNOT COPY)

Fill in the blank with the correct term. Some of the given choices will not be used. Descartes' rule of signs a vertical asymptote the leading-term test \(\quad\) an oblique the intermediate \(\quad\) asymptote value theorem direct variation the fundamental \(\quad\) inverse variation theorem of algebra a horizontal line a polynomial function a vertical line a rational function \(\quad\) parallel a one-to-one function \(\quad\) perpendicular a constant function a horizontal asymptote Two lines with slopes \(m_{1}\) and \(m_{2}\) are if and only if the product of their slopes is \(-1\)
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