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Chapter 9: Problem 18

Solve each inequality. Check your solutions. \(\log _{16} x \geq \frac{1}{4}\)

Short Answer

Expert verified
The solution is \( x \geq 2 \).

Step by step solution

01

Rewrite the Inequality Using Exponential Form

The inequality involves a logarithm, so let's convert it to its exponential form. The expression \( \log_{16} x \geq \frac{1}{4} \) can be rewritten as an exponential inequality: \( x \geq 16^{\frac{1}{4}} \).
02

Simplify the Exponential Expression

Now, simplify \( 16^{\frac{1}{4}} \). The base 16 can be rewritten as \( 2^4 \), so \( 16^{\frac{1}{4}} = (2^4)^{\frac{1}{4}} = 2^{4 \cdot \frac{1}{4}} = 2^1 = 2 \). Thus, we have \( x \geq 2 \).
03

Check the Solution

To verify our solution of \( x \geq 2 \), substitute \( x = 2 \) into the original inequality: \( \log_{16} 2\). Calculate \( \log_{16} 2 \) using the change of base formula: \[ \log_{16} 2 = \frac{\log_{10} 2}{\log_{10} 16} \].\Approximating using calculator values, \( \log_{10} 2 \approx 0.3010 \) and \( \log_{10} 16 \approx 1.2041 \), giving \[ \log_{16} 2 \approx \frac{0.3010}{1.2041} \approx 0.25 \]. This confirms that \( 0.25 \geq \frac{1}{4} \) is true, so \( x = 2 \) satisfies the inequality.

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Key Concepts

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

Logarithms
Logarithms are mathematical operations that help us solve equations involving exponential growth or decay. They are the inverse of exponentiation.
In simpler terms, if you know the result of a power and the base, a logarithm helps you determine the exponent.
Here are some important points about logarithms:
  • The logarithm of a number is the exponent to which the base must be raised to get that number. If \( b^y = x \), then \( \log_b(x) = y \).
  • Common bases for logarithms include 10 (common logarithm) and \( e \) (natural logarithm).
  • In the inequality \( \log_{16}(x) \geq \frac{1}{4} \), we want to find out when the logarithm of \( x \) to the base 16 is greater than or equal to 1/4.
Understanding logarithms is crucial, as they allow us to simplify complex calculations and solve for variables in exponential equations.
Exponential Expressions
Exponential expressions involve numbers raised to a power or exponent. They are used in various fields to model growth or decay phenomena.
  • An expression like \( a^b \) consists of a base \( a \) and an exponent \( b \).
  • The base indicates the number that is being multiplied, and the exponent tells you how many times to multiply the base by itself.
In the exercise, we converted a logarithmic inequality into an exponential form: \( x \geq 16^{\frac{1}{4}} \).
This step simplified the expression to make it easier to solve.
By changing \( 16 \) into \( 2^4 \), the expression \( 16^{\frac{1}{4}} \) became \( 2^{1} \).
Therefore, the inequality became \( x \geq 2 \), making the solution straightforward.
Change of Base Formula
The Change of Base Formula is a handy tool for evaluating logarithms that are not based on common numbers like 10 or \( e \).
It allows us to rewrite a logarithm in terms of logarithms with a different base, typically base 10, for easier computation.
  • The formula is \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \), where \( k \) is a new base of choice.
In our exercise, we needed the value of \( \log_{16}(2) \). By applying the change of base formula, it was computed as \[ \log_{16}(2) = \frac{\log_{10}(2)}{\log_{10}(16)} \].
This conversion made the calculation more straightforward, especially with a calculator.
Understanding and using the Change of Base Formula simplifies problems involving non-standard bases.

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Most popular questions from this chapter

For Exercises 53 and \(54,\) use the following information. The musical cent is a unit in a logarithmic scale of relative pitch or intervals. One octave is equal to 1200 cents. The formula to determine the difference in cents between two notes with frequencies \(a\) and \(b\) is \(n=1200\left(\log _{2} \frac{a}{b}\right)\). If the interval is 55 cents and the beginning frequency is 225 \(\mathrm{Hz}\) , find the final frequency.

In \(1928,\) when the high jump was first introduced as a women's sport at the Olympic Games, the winning women's jump was 62.5 inches, while the winning men's jump was 76.5 inches. Since then, the winning jump for women has increased by about 0.38\(\%\) per year, while the winning jump for men has increased at a slower rate, 0.3\(\%\) . If these rates continue, when will the women's winning high jump be higher than the men's?

For Exercises 55 and 56 , use the following information. If you deposit \(P\) dollars into a bank account paying an annual interest rate \(r\) (expressed as a decimal), with \(n\) interest payments each year, the amount \(A\) you would have after \(t\) years is \(A=P\left(1+\frac{r}{n}\right)^{n t} .\) Marta places \(\$ 100\) in a savings account earning 2\(\%\) annual interest, compounded quarterly. How long will it take for Marta's money to double?

Express each logarithm in terms of common logarithms. Then approximate its value to four decimal places. \(\log _{50} 23\)

Solve each equation or inequality. Check your solutions. $$ 3^{5 n+3}=3^{33} $$
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