If we move the decimal point to the left, we are making a number smaller, so we must multiply by a positive power of 10 to compensate. If we move the decimal point to the right, we must multiply by a negative power of 10.
ExampleA.6.13.
Write each number in scientific notation.
\(\displaystyle 62,000,000\)
\(\displaystyle 0.000431\)
Solution.
First, we position the decimal point so that there is just one nonzero digit to the left of the decimal.
To recover \(62,000,000\) from \(6.2\text{,}\) we must move the decimal point seven places to the right. Therefore, we multiply \(6.2\) by \(10^7\text{.}\)
To recover \(0.000431\) from \(4.31\text{,}\) we must move the decimal point seven places to the right. Therefore, we multiply \(4.31\) by \(10^{-4}\text{.}\)
Rational numbers are the integers and common fractions; we can represent them precisely in decimal form. But the best we can do for an irrational number is to write an approximate decimal form by rounding.
ExampleA.6.21.
Identify each number as rational or irrational.
\(\displaystyle \sqrt{6}\)
\(\displaystyle \dfrac{-5}{3}\)
\(\displaystyle \sqrt{16}\)
\(\displaystyle \sqrt{\dfrac{5}{9}}\)
Solution.
Irrational: \(~\sqrt{6}~\) is not the quotient of two integers.
Rational: \(~\dfrac{-5}{3}~\) is the quotient of two integers.
Rational: \(~\sqrt{16} = 4~\) is an integer.
Irrational: \(~\sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{3}~\text{,}\) but \(\sqrt{5}\) is irrational.
ExampleA.6.22.
Give a decimal approximation rounded to thousandths.
\(\displaystyle 5\sqrt{3}\)
\(\displaystyle \dfrac{-2}{3}\sqrt{21}\)
\(\displaystyle 2+\sqrt[3]{5}\)
Solution.
Use a calculator to evaluate.
Enter \(~5~ \boxed{\sqrt{}}~3\)ENTER and round to three decimal places: \(~8.660 \)
Enter (-)\(2\)\(\boxed{\sqrt{}}~\)\(21\))÷\(3\)ENTER and round to three decimal places: \(~-3.055\)
Enter \(2\)+MATH4\(5\)ENTER and round to three decimal places: \(~3.710\)
CheckpointA.6.23.
Identify each number as rational or irrational.
\(\displaystyle \sqrt{250}\)
\(\displaystyle \dfrac{\sqrt{3}}{2}\)
\(\displaystyle \dfrac{\sqrt{81}}{4}\)
\(\displaystyle \sqrt[3]{16}\)
Answer.
Irrational
Irrational
Rational
Irrational
CheckpointA.6.24.
Give a decimal approximation rounded to thousandths.
\(\displaystyle -6\sqrt[3]{5}\)
\(\displaystyle \dfrac{3}{5}\sqrt{76}\)
\(\displaystyle 7-\sqrt{19}\)
Answer.
\(\displaystyle -10.260\)
\(\displaystyle 5.231\)
\(\displaystyle 2.641\)
SubsubsectionA.6.2.3Using exponents and roots
Be careful to avoid tempting but false operations with exponents and roots.
ExampleA.6.25.
Which equation is a correct application of the laws of exponents?
\(20(1+r)^4 = 20+20r^4~~~~~\) or \(~~~~~(ab^t)^3 = a^3b^{3t}\)
\(2^{t/5} = (2^{1/5})^t~~~~~\) or \(~~~~~6.8(10)^t = 68^t\)
Solution.
The first statement is not correct. There is no law that says \((a+b)^n\) is equivalent to \(a^n+b^n\text{,}\) so \((1+r)^4\) is not equivalent to \(1^4+r^4\) or \(1+r^4\text{.}\)
However, it is true that \((ab)^n = a^nb^n\text{,}\) so in particular the second statement is true:
Not correct, but \(~N(0.94)^{-8.3} = \dfrac{N}{(0.94)^{8.3}}\)
Properties of Radicals.
Product Rule
\begin{equation*}
\sqrt[n]{ab}=\sqrt[n]{a}~\sqrt[n]{b}~~~~~~~\text{for } a, b \ge 0
\end{equation*}
Quotient Rule
\begin{equation*}
\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}~~~~~~~\text{for } a\ge 0,~~ b \gt 0
\end{equation*}
In general, it is not true that \(\sqrt[n]{a+b}\) is equivalent to \(\sqrt[n]{a}+\sqrt[n]{b}\text{,}\) or that \(\sqrt[n]{a-b}\) is equivalent to \(\sqrt[n]{a}-\sqrt[n]{b}\text{.}\)
ExampleA.6.30.
Which equation is a correct application of the properties of radicals?
\(\sqrt{x^4+81} = x^2 + 9~~~~~\) or \(~~~~~\sqrt[3]{P^2}~\sqrt[3]{1+r} = \sqrt[3]{P^2(1+r)}\)
\(\dfrac{\sqrt{x+y}}{\sqrt{x}} = \sqrt{y}~~~~~\) or \(~~~~~\dfrac{x+y}{\sqrt{x+y}} = \sqrt{x+y}\)
Solution.
The first statement is incorrect. There is no property that says \(~\sqrt[n]{a+b} = \sqrt[n]{a} + \sqrt[n]{b}\text{.}\)
However, it is true that \(~\sqrt[n]{a}~ \sqrt[n]{b} = \sqrt[n]{ab}~\text{,}\) so the second statement is correct.
The first statement is incorrect, because \(\dfrac{x+y}{x}\) is not equivalent to \(y\text{.}\)
The second statement is correct, because \(~\sqrt{x+y}~\sqrt{x+y} = x+y\text{.}\)
Decide whether each equation is a correct application of the properties of radicals. Write a correct statement if possible.
SubsubsectionA.6.3.2Convert between fractions and decimals
Rational exponents may also be written in decimal form.
ExampleA.6.41.
Convert \(~0.016~\) to a common fraction.
Solution.
The numerator of the fraction is 016, or 16. The last digit, 6, is in the thousandths place, so the denominator of the fraction is 1000. Thus, \(0.016=\dfrac{16}{1000}\text{.}\) We can reduce this fraction by dividing top and bottom by 8:
Convert \(~\dfrac{4}{15}~\) to a decimal fraction.
Answer.
\(0.2\overline{6}\)
SubsubsectionA.6.3.3Solve equations
To solve an equation of the form \(~x^n = k\text{,}\) we can raise both sides to the reciprocal of the exponent:
\begin{align*}
(x^n)^{1/n} \amp = k^{1/n}\\
x \amp = k^{1/n}
\end{align*}
because \(~(x^n)^{1/n} = x^{n(1/n)} = x^1\text{.}\)
ExampleA.6.47.
Solve \(~0.6x^4 = 578\text{.}\) Round your answer to hundredths.
Solution.
First, we isolate the power.
\begin{align*}
0.6x^4 \amp = 578 \amp\amp \blert{\text{Divide both sides by 0.6.}}\\
x^4 \amp = 963.\overline{3}
\end{align*}
We raise both sides to the reciprocal of the power.
\begin{align*}
(x^4)^{1/4} \amp = (963.\overline{3})^{1/4} \amp\amp \blert{\text{By the third law of exponents,}~ (x^4)^{1/4}=x.}\\
x \amp = 5.57
\end{align*}
To evaluate \((963.\overline{3})^{1/4}\text{,}\) enter \(~~\text{ANS}\)^\(.25\)ENTER
ExampleA.6.48.
Solve \(~x^{2/3}-4=60\text{.}\)
Solution.
First, we isolate the power.
\begin{align*}
x^{2/3}-4 \amp = 60 \amp\amp \blert{\text{Add 4 to both sides.}}\\
x^{2/3} \amp = 64
\end{align*}
We raise both sides to the reciprocal of the power.
Or we can evaluate \(~64^{3/2}~\) by entering \(~~64\)^\(1.5\)ENTER
ExampleA.6.49.
Solve \(~18x^{0.24} = 6.5\text{.}\) Round your answer to thousandths.
Solution.
First, we isolate the power.
\begin{align*}
18x^{0.24} \amp = 6.5 \amp\amp \blert{\text{Divide both sides by 18.}}\\
x^{0.24} \amp = 0.36\overline{1}
\end{align*}
We raise both sides to the reciprocal of the power.
\begin{align*}
\left(x^{0.24}\right)^{1/0.24} \amp = (0.36\overline{1})^{1/0.24}\\
x \amp = 0.014
\end{align*}
We evaluate \((0.36\overline{1})^{1/0.24}\) by entering \(~~\text{ANS}\)^(\(1\)÷\(.24\))ENTER
CheckpointA.6.50.
Solve \(~4x^5 = 1825~\text{.}\) Round your answer to thousandths.
Answer.
\(3.403\)
CheckpointA.6.51.
Solve \(~\dfrac{3}{4}x^{3/4} = 36~\text{.}\) Round your answer to thousandths.
Answer.
\(174.444\)
CheckpointA.6.52.
Solve \(~0.2x^{1.4}+1.8=12.3~\text{.}\) Round your answer to thousandths.
Answer.
\(16.931\)
SubsectionA.6.4Working with Radicals
SubsubsectionA.6.4.1Perfect squares
To simplify a radical, we factor out the largest perfect square.
ExampleA.6.53.
Find the missiing factor.
\(60x^9 = 3x^3 \cdot~\) ?
\(9x^3 - 3x^9 \cdot~\) ?
Solution.
We mentally divide \(60x^9\) by \(3x^3\) to find \(\dfrac{60x^9}{3x^3} = 20x^6\text{.}\) The missing factor is \(20x^6\text{.}\)
We mentally divide each term by \(3x^3\) to find \(\dfrac{9x^3}{3x^3} = 3\) and \(\dfrac{3x^9}{3x^3}=x^6\text{.}\) The missing factor is \(3-x^6\text{.}\)
ExampleA.6.54.
Factor out the largest perfect square.
\(\displaystyle 108a^5b^2\)
\(\displaystyle \dfrac{a^2+4a^4}{8}\)
Solution.
By trial and error, we find that 36 is the largest square that divides 108. From each power, we can factor out the power with the largest possible even exponent, namely \(a^4\) and \(b^2\text{.}\) Thus, we factor out \(36a^4b^2\) to find \(108a^5b^2 =36a^4b^2 \cdot 3a\text{.}\)
The largest even power that divides into both \(a^2\) and \(a^4\) is \(a^2\text{,}\) so we factor \(a^2\) from the numerator:
No: we cannot combine radicals with addition or subtraction. You can check that \(\sqrt{6} \approx 2.4495\text{,}\) but \(\sqrt{2}+\sqrt{4} \approx 1.4142+2=3.4142\text{.}\)
The largest perfect square that divides 20 is 4. We write the radicand as the product of two factors, one containing the perfect square and the largest possible even powers of the variables. That is,
Finally, we add 4 to both sides to find the solution, \(x=31\text{.}\) We do not have to check for extraneous solutions when we cube both sides of an equation, but it is a good idea to check the solution for accuracy anyway.
Check: We substitute \(\alert{31}\) for \(x\) into the left side of the equation.
Explain why \(\sqrt{x^2}=x\) is not true for all values of \(x\text{.}\)
Solution.
Recall that the symbol \(\sqrt{a}\) means the non-negative square root of \(a\text{.}\) If \(x\) is a negative number, for example \(x=-6\text{,}\) then \(x^2=(-6)^2=36\text{,}\) and not \(\sqrt{x^2}=\sqrt{36}=6\text{,}\) not \(-6\text{.}\) So if \(x\) is a negative number, \(\sqrt{x^2}\not= x\) In fact, \(\sqrt{x^2}= \abs{x}\text{.}\)
ExampleA.6.80.
For what values of \(x\) is \(~\sqrt{(x-5)^2} = x-5~\text{?}\)
Solution.
\(~\sqrt{(x-5)^2} = x-5~\) when \(x-5\) is positive or zero, that is for \(x \ge 5\text{.}\) If \(x \lt 5\text{,}\) then \(x-5\) is negative. But the \(\sqrt{}\) symbol returns only the positive root, so we use absolute value bars to indicate that the root is positive:
