Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 6.2
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Exercise 99 Page 284

Practice makes perfect
a

Let's name the number we need to multiply 8 by to get 1 as n. We want to know what we should multiply n by to get 1. We can write the following equation.

8n=1 By solving this equation for n, we can determine what this number is.

8n=1
n=1/8

b

Like in Part A, we will write an equation containing the product of x and our number n on one side and 1 on the other side.

nx=1 By solving this equation for n, we can determine what this number is.

nx=1
n=1/x

c

Examining the equation, we see that m is raised to the power of 8. To solve for m we have to eliminate the exponent. We can do that by taking the 8^(th) root of both sides. Notice that because the exponent is even we get two solutions.

sqrt(m^8)=± sqrt(40) ⇕ m=± sqrt(40) However many, if not all, calculators do not have the 8^(th) root as a button. Instead, we can raise both sides to the inverse of 8, which is the equivalent to taking the 8^(th) root. (m^8)^(18)=± (40)^(18) ⇕ m=± (40)^(18) Using a calculator, we can determine the right-hand side.

As we can see, m≈ ± 1.5858.... We can check it by raising the two solutions to the power of 8 on our calculator.

d

From Part C, we know that to eliminate an exponent we have to raise the power to the inverse of the exponent. Additionally, if the exponent is even, we get a positive and a negative solution.

(n^6)^(16)=± (300)^(16) ⇕ m=± (300)^(16) Again, we will use the calculator to solve this.

We get the solutions n ≈ ± 2.59.

e

As we talked about in previous parts, we have to raise a power to the inverse of its exponent to solve for x. Additionally, if a is even, we get a positive and a negative solution.

a is even:& x^a=b x=± b^(1a) a is odd:& x^a=b x= b^(1a)