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Chapter 4: Problem 47

Rewrite each expression as simply as you can. $$m^{7} \cdot m^{-5} \cdot b^{5}$$

Short Answer

Expert verified
m^2 \cdot b^5

Step by step solution

01

- Apply Rules of Exponents to Same Bases

Combine the terms with the same base using the rule of exponents which states that when you multiply like bases, you add the exponents. For the terms with base m, add the exponents: \(m^7 \cdot m^{-5}\).
02

- Simplify the Combined Exponents of m

Add the exponents of m: \(7 + (-5) = 2\). So, \(m^7 \cdot m^{-5} = m^2\).
03

- Combine All Simplified Terms

Combine the simplified terms: \(m^2\) and \(b^5\). Therefore, the expression simplifies to \(m^2 \cdot b^5\).

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Key Concepts

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

rules of exponents
When dealing with exponents, it's important to understand their rules. The primary rule used in this exercise is: When you multiply terms that have the same base, you add their exponents. Here鈥檚 how it works: If you have an expression like \(a^m \cdot a^n\), you can simplify it to \(a^{(m+n)}\).

The same goes when you have negative exponents. For instance, in our exercise, \(m^7 \cdot m^{-5}\), the base is the same (m), so you add the exponents: 7 and -5. This results in \(m^2\).

Here鈥檚 a quick tip:
  • If you add a positive and a negative exponent, think of it as a normal subtraction.
  • If the sum of the exponents is zero, the expression equals 1, since any number to the zero power is 1.
combining like terms
Another essential concept in simplifying expressions is combining like terms. Like terms are terms that have the same variable raised to the same power.

For example:
  • \(3a + 2a\), here both terms are like terms because they have the same variable 'a', and when you add them, you get \(5a\).
  • In contrast, \(3a + 2b\) are not like terms because they have different variables.
In our exercise, once we simplified the exponents of 'm', we combined \(m^2\) and \(b^5\) because they are not like terms. We cannot simplify these further since they are distinct variables.

Always remember to combine only terms that have the exact same variables raised to the same powers.
simplifying expressions
Simplifying expressions involves making them as straightforward as possible. This often requires applying several algebraic rules.

Let鈥檚 summarize the steps:
  • Identify and combine like terms.
  • Use the rules of exponents to handle terms with the same base.
  • Perform any addition, subtraction, multiplication, or division needed to further simplify.
In our exercise, we combined the same bases using the rules of exponents, then simplified the resulting exponents, and finally combined all remaining terms to reach the simplest form. The goal is always to make the expression shorter and easier to understand without changing its value. In our example, the simplified form is \(m^2 \cdot b^5\).

Once you grasp these concepts鈥攔ules of exponents, combining like terms, and simplifying expressions鈥攜ou'll find algebra much easier and more intuitive.

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Most popular questions from this chapter

Examine this table. a. Use the table to estimate the solutions of \(t(t-3)=5\) to the nearest integer. b. If you were searching for solutions by making a table with a calculator, what would you have to do to find solutions to the nearest tenth? c. Find two solutions of \(t(t-3)=5\) to the nearest tenth. $$ \begin{array}{rr}{t} & {t(t-3)} \\ {-2} & {10} \\ {-1} & {4} \\ {0} & {0} \\\ {1} & {-2} \\ {2} & {-2} \\ {3} & {0} \\ {4} & {4} \\ {5} & {10} \\\ {6} & {18} \\ {7} & {28}\end{array} $$

Solve the equation \((x+2)(x-3)=14\) by constructing a table of values. Use integer values of \(x\) between \(-6\) and \(6 .\)

For each table, tell whether the relationship between x and y could be linear, quadratic, or an inverse variation, and write an equation for the relationship. $$ \begin{array}{|c|c|c|c|c|c|}\hline x & {0.5} & {2} & {4} & {5} & {20} \\\ \hline y & {5} & {1.25} & {0.625} & {0.5} & {0.125} \\ \hline\end{array} $$

Economics Andre bought 11 books at a used book sale. Some cost 25\(\phi\) each; the others cost 35\(鈧琝) each. Andre spent \(\$ 3.15 .\) a. Write a system of two equations to describe this situation, and solve it by substitution. b. How many books did Andre buy for each price?

Recall that for linear equations, first differences are constant; and that for quadratic equations, second differences are constant. Determine whether the relationship in each table could be linear, quadratic, or neither. $$ \begin{array}{|c|c|}\hline x & {y} \\ \hline-3 & {0} \\ {-2} & {-0.5} \\\ {-1} & {-1} \\ {0} & {-1.5} \\ {1} & {-2} \\ {2} & {-2.5} \\\ \hline\end{array} $$
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