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Chapter 1: Problem 5

Simplify the following expressions, giving the answers with positive indices and without brackets: (a) \(2^{3} \times 2^{-4}\) (b) \(2^{3} \div 2^{-4}\) (c) \(\left(2^{3}\right)^{-4}\) (d) \(3^{1 / 3} \times 3^{3 / 3}\) (e) \((36)^{-1 / 2}\) (f) \(16^{3 / 4}\)

Short Answer

Expert verified
(a) \(\frac{1}{2}\); (b) \(2^7\); (c) \(\frac{1}{2^{12}}\); (d) \(3^{4/3}\); (e) \(\frac{1}{6}\); (f) \(8\).

Step by step solution

01

Apply exponent rules to simplify (a)

When multiplying powers with the same base, we add the exponents. So for \(2^{3} \times 2^{-4}\), we can express it as:\[ 2^{3 + (-4)} = 2^{-1} \]Since positive indices are required, we rewrite \(2^{-1}\) as the reciprocal:\[ \frac{1}{2} \]
02

Simplify (b) using division rule

When dividing powers with the same base, subtract the exponents. So for \(2^{3} \div 2^{-4}\), we have:\[ 2^{3 - (-4)} = 2^{7} \]
03

Use negative power rule for (c)

For the power of a power, multiply the exponents: \(\left(2^{3}\right)^{-4}\) becomes:\[ 2^{3 \times (-4)} = 2^{-12} \]Convert to a positive index:\[ \frac{1}{2^{12}} \]
04

Simplify (d) using fractional exponents

Add the exponents for the multiplication of terms with the same base. Thus, \(3^{1/3} \times 3^{3/3}\) becomes:\[ 3^{1/3 + 1} = 3^{4/3} \]
05

Solve (e) using the negative and fractional power rule

Rewrite the expression \((36)^{-1/2}\) as a reciprocal of the square root:\[ \frac{1}{\sqrt{36}} = \frac{1}{6} \]
06

Simplify (f) using fractional exponent rules

Using the rule \(a^{m/n} = \sqrt[n]{a^m}\), simplify \(16^{3/4}\) as:\[ (16^{1/4})^3 = (2)^3 = 8 \]

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Key Concepts

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

Multiplication of Exponents
When multiplying exponents with the same base, such as in \(2^3 \times 2^{-4}\), the rule is to add the exponents together. This concept might seem a bit mysterious at first, but let鈥檚 break it down:
  • Think of it like multiplying numbers with a base or foundation that stays the same.
  • You simply look at the exponents and add them up.
In our case, the expression becomes \(2^{3 + (-4)} = 2^{-1}\). This shows us that multiplying is essentially accumulating powers. When the sum of the exponents is negative, it means the base moves to the denominator position as a reciprocal. So, \(2^{-1}\) becomes \(\frac{1}{2}\). This transformation helps convert negative indices into positive ones, which is often a requirement in final answers.
Division of Exponents
Division of exponents functions a bit differently from multiplication. Here, you subtract the exponents instead of adding them. For example, consider \(2^3 \div 2^{-4}\):
  • This operation asks us to perform \(2^{3 - (-4)}\).
  • This translates to adding the opposite exponent, which results in \(3 + 4 = 7\).
Thus, the expression simplifies to \(2^7\). This rule is akin to dividing repeated multiplications, so the focus stays on subtracting exponents. The idea of switching the signs during subtraction can be a bit tricky, but with practice, it becomes much clearer.
Negative Exponents
Negative exponents indicate that the base should be taken as the reciprocal. For example, if you come across \(2^{-12}\), as in \((2^3)^{-4} = 2^{-12}\), it means:
  • You flip the base to calculate \(\frac{1}{2^{12}}\).
  • Recognizing this flip is key to understanding negative exponents.
This property is indeed one of the most useful in mathematics because it simplifies the handling of very large or very small numbers. Anytime you see a negative exponent, it helps to remember that it鈥檚 a call to action to invert the base.
Fractional Exponents
Fractional exponents bring a geometric understanding to exponents by involving roots. Let鈥檚 take \(16^{3/4}\) as an example:
  • Here, \(16^{3/4}\) indicates finding the fourth root of 16, then raising the result to the power of 3.
  • This operation becomes equivalent to \( (2)^3 = 8 \) since \(16^{1/4} = 2\).
Another interesting case is \(3^{1/3} \times 3^{3/3}\), where we add the exponents, resulting in \(3^{4/3}\). Fractional exponents can initially be confusing, but they simplify many operations and problems, especially when dealing with roots. Remember the rule: \(a^{m/n}\) is the same as taking the \(n\)-th root of \(a\) raised to the \(m\)-th power.

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Most popular questions from this chapter

Given \(a_{0}=2, a_{1}=-1, a_{2}=-4, a_{3}=5, a_{4}=3\) and \(b_{0}=1, b_{1}=1, b_{2}=2, b_{3}=-1, b_{4}=2\) calculate (a) \(\sum_{k=0}^{4} a_{i}\) (b) \(\sum_{i=1}^{3} a_{i}\) (c) \(\sum_{k=1}^{2} a_{k} b_{k}\) (d) \(\sum_{j=0}^{4} b_{j}^{2}\)

Let \(x\) be a constant such that \(|x|<1\). Find the solution of $$ T_{n+2}-2 x T_{n+1}+T_{n}=0, \quad T_{0}=1, \quad T_{1}=x $$Find \(T_{2}, T_{3}\) and \(T_{4}\) also directly by recursion and deduce that \(\cos \left(2 \cos ^{-1} x\right)=2 x^{2}-1\) and express \(\cos \left(3 \cos ^{-1} x\right)\) and \(\cos \left(4 \cos ^{-1} x\right)\) as polynomials in \(x\)

Rearrange the following quadratic expressions by completing the square. (a) \(x^{2}+x-12\) (b) \(3-2 x+x^{2}\) (c) \((x-1)^{2}-(2 x-3)^{2}\) (d) \(1+4 x-x^{2}\)

Consider the following puzzle: how many single, loose, smooth \(30 \mathrm{~cm}\) bricks are necessary to form a single leaning pile with no part of the bottom brick under the top brick? Begin by considering. a pile of 2 bricks. The top brick cannot project further than \(15 \mathrm{~cm}\) without collapse. Then consider a pile of 3 bricks. Show that the top one cannot project further than \(15 \mathrm{~cm}\) beyond the second one and that the second one cannot project further than \(7.5 \mathrm{~cm}\) beyond the bottom brick (so that the maximum total lean is \(\left.\left(\frac{1}{2}+\frac{1}{4}\right) 30 \mathrm{~cm}\right)\). Show that the maximum total lean for a pile of 4 bricks is \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right) 30 \mathrm{~cm}\) and deduce that for a pile of \(n\) bricks it is \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2 n+2}\right) 30 \mathrm{~cm}\). Hence solve the puzzle.

Find the general solution of the linear recurrence relation $$ (n+1)^{2} x_{n+1}-n^{2} x_{n}=1, \quad \text { for } n \geqslant 1 $$ (Hint: The coefficients are not constants. Use the substitution \(z_{n}=n^{2} x_{n}\) to find a constant-coefficient equation for \(z_{n^{-}}\). Find the general solution for \(z_{n}\) and hence for \(x_{n}\).)
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