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Chapter 4: Problem 6

Fill in the blank to make a true statement. $$ 4^{\square}=\frac{1}{64} $$

Short Answer

Expert verified
The blank should be filled with \-3\.

Step by step solution

01

Recognize the relationship between exponents and fractions

When a base raised to a power results in a fraction, it usually means the exponent is negative. This is because a negative exponent indicates the reciprocal of the base raised to the positive of that power.
02

Express \(\frac{1}{64}\) as a power of 4

To find the exponent, determine the power of 4 that equals 64. We know that \({{4}^{3} = 64}\). Therefore, \({\frac{1}{64} = \frac{1}{{4}^3} = {4}^{-3}}\).
03

Write the final equation

By expressing \(\frac{1}{64}\) as \({4}^{-3}\), we can fill in the blank. Thus, \({{4}^{\text{-3}} = \frac{1}{64}}\).

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Key Concepts

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

Negative Exponents
Negative exponents can be a bit tricky at first. However, they're just inverses of positive exponents. If you see a base raised to a negative exponent, like \(a^{-n}\), it means you're looking at the reciprocal of the base raised to the positive exponent. So, \(a^{-n} = \frac{1}{a^n}\).

Take our given problem, for example. We need to fill in the blank for \(4^{\text{_____}} = \frac{1}{64}\). Since \[ 4^3 = 64 \], the reciprocal of that is \[ \frac{1}{64} = \frac{1}{4^3} \], which simplifies to \[ 4^{-3}\].

By understanding negative exponents, we can easily flip the fraction and find the missing exponent.
Reciprocals
Reciprocals are like flipping a number. If you have a fraction, you switch its numerator and denominator. For example, the reciprocal of \(a\) is \(\frac{1}{a}\) and vice versa.

In our problem, \( \frac{1}{64} \) is given. Recognizing that \( \frac{1}{64} \) is the reciprocal of \( 64 \), we can write it as a reciprocating power of 4.

Since \[4^3 = 64 \], the reciprocal is \[ \frac{1}{4^3} \], aligning it with our formula for negative exponents: \[ 4^{-3} \].

Understanding reciprocals helps break down problems involving fractions and exponents.
Power of a Base
The power of a base means raising a number to an exponent. The base is the number you multiply by itself, and the exponent tells you how many times to use that base.

In our case, 4 is the base, and we need to find which power of 4 equals \( \frac{1}{64} \). Knowing the relationship between bases and exponents is crucial.

First, find \[ 4^3 = 64 \]. Then, recognizing the reciprocal \[ \frac{1}{64} \], rewrite it in terms of negative exponents: \[ \frac{1}{64} = \frac{1}{4^3} = 4^{-3} \].

Mastering the power of a base concept allows you to handle both positive and negative exponents confidently.

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Most popular questions from this chapter

Show that \(-\ln \left(x-\sqrt{x^{2}-1}\right)=\ln \left(x+\sqrt{x^{2}-1}\right)\).

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(6 \log _{5}(4 p-3)=18\)

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log \left(p^{2}+6 p\right)=\log 7\)

9\. The population of the United States \(P(t)\) (in millions) since January 1,1900 , can be approximated by $$P(t)=\frac{725}{1+8.295 e^{-0.0165 t}}$$ where \(t\) is the number of years since January \(1,1900 .\) (See Example 6\()\) a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the U.S. population on January \(1,2020 .\) Round to the nearest million. c. Use the function to predict the U.S. population on January 1,2050 . d. Determine the year during which the U.S. population will reach 500 million. e. What value will the term \(\frac{8.295}{e^{0.0165 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

Use the model \(A=P e^{r t} .\) The variable \(A\) represents the future value of \(P\) dollars invested at an interest rate \(r\) compounded continuously for \(t\) years. If a couple has \(\$ 80,000\) in a retirement account, how long will it take the money to grow to \(\$ 1,000,000\) if it grows by \(6 \%\) compounded continuously? Round to the nearest year.
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