QuanticaAlgebra IOpen in the course3.1 kept the larger exponent on top of every quotient, and that restriction is now gone. Once the bottom exponent can be the larger one, the quotient rule's subtraction runs past zero, so \(x^{0}\) and \(x^{-3}\) have to mean something. There is a second change. The base is a letter now, and a letter can be zero without looking like it.
Problem
A column lists powers of \(d\) for some nonzero \(d\), and each entry is the one above it divided by \(d\). Reading down, the entries are \(d\cdot d\cdot d\), then \(d\cdot d\), then \(d\). Take one more step down, which lands on \(d^{0}\). Enter its value.
Show a hint
- Each step down divides by \(d\), and the last step is no different from the ones above it.
- The entry above the one you want is \(d\) itself, so the step is \(d\div d\).
Show the full solution
Every step down the column divides by \(d\), including the last one. The entry above \(d^{0}\) is \(d\), so $$d^{0}=d\div d=\boxed{1}.$$ The base cancels itself. Nothing about the step changed just because the exponent reached zero.
Problem
A table lists the value of \(t^{0}\) for \(t=13\), \(t=-9\), \(t=0\) and \(t=50\). Check each value of \(t\) against the zero exponent rule before you count. For how many of the four is \(t^{0}\) equal to \(1\)?
Show a hint
- The zero power needs a base that is not zero. Check each value of \(t\) against that.
- A negative base is still a nonzero base, so \((-9)^{0}=1\). Now test \(t=0\) against the condition.
Show the full solution
\(13^{0}\), \((-9)^{0}\) and \(50^{0}\) are all \(1\), and \(0^{0}\) has no value, so \(\boxed{3}\) of the four are equal to \(1\). The rule \(a^{0}=1\) needs a nonzero base, and a negative base still counts as nonzero. Zero is left out because \(t^{0}\) comes from \(\frac{t^{n}}{t^{n}}\), which divides by zero when \(t=0\).
The column is one route to \(x^{0}\), and the quotient rule is the other. It reaches further than the column does, because subtracting exponents does not stop at zero. The next two problems run one fraction through both readings and then make the readings agree.
Problem
Simplify \(\frac{h^{2}}{h^{8}}\) for nonzero \(h\) by writing out the factors and cancelling. Nothing here needs the quotient rule. Enter what remains, typed like \(1/y^5\).
Show a hint
- Two factors on top, eight on the bottom. Pair them off.
- Each factor on top cancels one on the bottom. Count how many factors are left underneath.
Show the full solution
The two factors of \(h\) on top cancel two of the eight underneath, which leaves \(8-2=6\) factors on the bottom and nothing but \(1\) on top. $$\frac{h^{2}}{h^{8}}=\boxed{\frac{1}{h^{6}}}$$
Problem
Now read the same fraction \(\frac{h^{2}}{h^{8}}\) through the quotient rule from 3.1, which subtracts the exponents. Nothing in that rule ever said the top exponent had to be the larger one. Enter the exponent it puts on \(h\).
Show a hint
- The rule is \(\frac{x^{m}}{x^{n}}=x^{m-n}\). Put \(2\) and \(8\) in and do the subtraction.
- Subtract in the order the rule gives, so it is \(2-8\) and not \(8-2\). A negative result is fine as an exponent.
Show the full solution
The quotient rule subtracts the exponents whether or not the top one is larger. $$\frac{h^{2}}{h^{8}}=h^{2-8}=h^{-6}$$ so the exponent is \(\boxed{-6}\). The last problem read the same fraction as \(\frac{1}{h^{6}}\), and one fraction has one value, so \(h^{-6}\) and \(\frac{1}{h^{6}}\) are the same thing.
Problem
Rewrite \(v^{-4}\) with a positive exponent, then evaluate it at \(v=3\). Enter the value as a fraction in lowest terms.
Show a hint
- A negative exponent puts the matching positive power under a \(1\).
- So this is \(\frac{1}{v^{4}}\). Now put \(3\) in for \(v\).
Show the full solution
A negative exponent is a reciprocal, so \(v^{-4}=\frac{1}{v^{4}}\). At \(v=3\) that is $$\frac{1}{3^{4}}=\boxed{\frac{1}{81}}.$$ A negative exponent puts the power under the bar and never turns the value negative, so \(-81\) and \(-\frac{1}{81}\) are both wrong turns.
Problem
Simplify \(\frac{1}{g^{-7}}\) for nonzero \(g\), leaving no fraction bar and no negative exponent. Enter the result, typed like \(y^5\).
Show a hint
- Rewrite the bottom first. What does \(g^{-7}\) equal with a positive exponent?
- The bottom becomes \(\frac{1}{g^{7}}\), and dividing by a fraction multiplies by its reciprocal.
Show the full solution
The bottom is \(g^{-7}=\frac{1}{g^{7}}\), so the whole thing is \(1\) divided by \(\frac{1}{g^{7}}\), and dividing by a fraction multiplies by its reciprocal. $$\frac{1}{g^{-7}}=\boxed{g^{7}}$$ A factor with a negative exponent on the bottom moves up top and its exponent turns positive. The move runs both ways.
A monomial has a number out front and powers of letters after it, and an exponent written on one of those letters has nothing to do with the number. This is where the most common wrong answer of the chapter lives. The next two problems are built to make the difference impossible to argue with.
Problem
A student writes \(5c^{-1}=\frac{1}{5c}\). Test the claim the way 1.4 did, by picking a number. Take \(c=2\) and work out what \(5c^{-1}\) is really worth there. Enter that value as a fraction in lowest terms or an exact decimal.
Show a hint
- The \(-1\) is written on \(c\). Ask what it is attached to before you move anything.
- \(5c^{-1}\) means \(5\cdot c^{-1}\), so the \(5\) is a separate factor sitting out front. Only \(c^{-1}\) becomes a reciprocal.
Show the full solution
The exponent is written on \(c\) alone, so \(5c^{-1}=5\cdot\frac{1}{c}=\frac{5}{c}\), and at \(c=2\) that is $$\boxed{\frac{5}{2}}.$$ The student's version gives \(\frac{1}{5\cdot 2}=\frac{1}{10}\), which is \(25\) times too small. The \(5\) never moves, because no exponent was ever written on it.
Problem
One pair of parentheses is the only difference between \(7q^{-2}\) and \((7q)^{-2}\), and they are not the same expression. Rewrite \((7q)^{-2}\) with no negative exponent. Enter the result, typed like \(1/(3y^4)\).
Show a hint
- In \(7q^{-2}\) the exponent is written on \(q\). In \((7q)^{-2}\) the parentheses put the whole product \(7q\) under it.
- So the second one is \(\frac{1}{(7q)^{2}}\), and squaring a product squares both factors, the way 3.1's power of a product did.
Show the full solution
The exponent applies to the whole product \(7q\), so $$(7q)^{-2}=\frac{1}{(7q)^{2}}=\boxed{\frac{1}{49q^{2}}}.$$ Without the parentheses, \(7q^{-2}=\frac{7}{q^{2}}\), which keeps the \(7\) upstairs. At \(q=2\), \((7q)^{-2}\) is \(\frac{1}{196}\) while \(7q^{-2}\) is \(\frac{7}{4}\), a factor of \(343\) apart.
Problem
Simplify \((4z^{-3})^{-2}\) so that no exponent is negative. The outer \(-2\) reaches the \(4\) as well as the \(z^{-3}\), the way 3.1's power of a product did. Enter the result, typed like \(y^5/9\).
Show a hint
- Handle the two factors inside separately. What is \(4^{-2}\), and what is \((z^{-3})^{-2}\)?
- Power of a power multiplies the exponents, and \((-3)\times(-2)\) is positive.
Show the full solution
The outer exponent applies to every factor inside the parentheses, so take the \(4\) and the \(z^{-3}\) one at a time. Power of a power multiplies exponents and \((-3)(-2)=6\), so \((z^{-3})^{-2}=z^{6}\), while the coefficient gives \(4^{-2}=\frac{1}{16}\). $$(4z^{-3})^{-2}=\boxed{\frac{z^{6}}{16}}$$ Leaving the \(4\) alone gives \(4z^{6}\), and flipping the wrong way gives \(\frac{16}{z^{6}}\).
Problem
Simplify \(\frac{12r^{3}}{18r^{10}}\) completely, with no negative exponent left. The coefficients and the powers of \(r\) do not get the same treatment. Enter the result, typed like \(3/(4y^2)\), with the whole denominator inside the parentheses.
Show a hint
- Split it into a number part and an \(r\) part. The numbers reduce, the exponents subtract.
- The exponents give \(3-10\), which is negative. A factor with a negative exponent moves to the denominator, and its exponent turns positive there.
Show the full solution
The coefficients reduce, \(\frac{12}{18}=\frac{2}{3}\), and the exponents subtract, \(3-10=-7\), so partway you have \(\frac{2}{3}r^{-7}\). $$\frac{2}{3}r^{-7}=\frac{2}{3}\cdot\frac{1}{r^{7}}=\boxed{\frac{2}{3r^{7}}}$$ Only the factor with the negative exponent moves to the denominator. The \(\frac{2}{3}\) has no negative exponent, so it stays where it is.
Problem
Simplify \(\frac{20a^{4}b^{6}}{45a^{9}b^{6}}\) for nonzero \(a\) and \(b\), leaving no negative exponent. One of the two letters disappears entirely, and the fact that it is nonzero is what lets that happen. Enter the result, typed like \(5/(7y^3)\).
Show a hint
- Three separate jobs. Reduce the numbers, subtract the \(a\) exponents, subtract the \(b\) exponents.
- The \(b\) exponents subtract to \(0\), and \(b^{0}=1\) because \(b\) is not zero. A factor of \(1\) can be dropped.
Show the full solution
Reduce the coefficients, \(\frac{20}{45}=\frac{4}{9}\). Subtract on \(a\), \(4-9=-5\). Subtract on \(b\), \(6-6=0\), and \(b^{0}=1\) since \(b\) is nonzero, so the \(b\) factors cancel and leave nothing behind. $$\frac{4}{9}a^{-5}=\boxed{\frac{4}{9a^{5}}}$$ Only \(a^{-5}\) moves to the denominator, since the \(4\) carries no exponent. Subtracting the other way gives \(a^{5}\) on top, which is the common miss.
Every exponent in this lesson was a whole number, positive, zero or negative, and the laws from 3.1 handled all of them without a single new rule. Nothing in those laws says an exponent has to be a whole number. 3.3 asks what \(x^{1/2}\) could mean, and the answer comes from the same place these did, whichever value keeps the laws working.
Practice these ideas
Practice
For nonzero \(k\), evaluate \(8k^{0}+(8k)^{0}\).
Show the solution
The first term is \(8k^{0}=8\cdot 1=8\) and the second is \((8k)^{0}=1\), so the sum is \(8+1=\boxed{9}\). Reading the first term as \((8k)^{0}\) instead gives \(1+1=2\), the common slip. There the exponent reaches only \(k\), so the \(8\) stays as a factor.
Practice
\(u^{0}=1\) holds for every value of \(u\) but one. Enter the value it fails for.
Show the solution
\(u^{0}=1\) fails at \(u=\boxed{0}\). Going from \(u^{1}\) down to \(u^{0}\) divides \(u\) by itself, and at \(u=0\) that is \(0\div 0\), which is undefined. Every other base works, negative ones included.
Practice
Rewrite \(j^{-9}\) with a positive exponent. Type it like \(1/y^4\).
Show the solution
A negative exponent means one over the matching positive power, so \(j^{-9}=\boxed{\dfrac{1}{j^{9}}}\). The minus sign moves the power to the bottom of a fraction. It does not make the value negative.
Practice
Evaluate \(w^{-3}\) at \(w=2\). Enter it as a fraction in lowest terms.
Show the solution
Rewriting gives \(w^{-3}=\frac{1}{w^{3}}\), and at \(w=2\) that is \(\frac{1}{2^{3}}=\boxed{\frac{1}{8}}\). Treating the minus as a sign on the value gives \(-8\), which is wrong in both size and sign.
Practice
Rewrite \(\frac{6}{f^{-2}}\) with no negative exponent. Type it like \(3y^4\).
Show the solution
The \(f^{-2}\) crosses the bar and the sign of its exponent flips, so \(\frac{6}{f^{-2}}=\boxed{6f^{2}}\). The \(6\) has no exponent written on it, so it does not move.
Practice
Simplify \(\frac{s^{5}}{s^{13}}\) with no negative exponent left. Type it like \(1/y^4\).
Show the solution
Dividing powers of the same base subtracts the exponents, and \(5-13=-8\), so the quotient is \(s^{-8}=\boxed{\frac{1}{s^{8}}}\). Cancelling gives the same result, since five of the thirteen factors in the denominator pair off with the five on top and eight stay below the bar.
Practice
Rewrite \(10p^{-3}\) with no negative exponent. Type it like \(2/y^5\).
Show the solution
The exponent sits on \(p\) alone, so only \(p^{-3}\) becomes a reciprocal and \(10p^{-3}=\boxed{\frac{10}{p^{3}}}\). Dragging the \(10\) down as well gives \(\frac{1}{10p^{3}}\), which is 100 times too small.
Practice
Evaluate \(15e^{-2}\) at \(e=3\). Enter it as a fraction in lowest terms.
Show the solution
The exponent is written on \(e\) alone, so \(15e^{-2}=\frac{15}{e^{2}}\), and at \(e=3\) that is \(\frac{15}{9}=\boxed{\frac{5}{3}}\). Moving the \(15\) down too gives \(\frac{1}{135}\), which is 225 times smaller.
Practice
Rewrite \((3b)^{-3}\) with no negative exponent. Simplify the number too. Type it like \(1/(2y^4)\).
Show the solution
\((3b)^{-3}=\frac{1}{(3b)^{3}}=\boxed{\frac{1}{27b^{3}}}\). Because of the parentheses the exponent applies to both factors, so the \(3\) gets cubed along with the \(b\). Without them \(3b^{-3}\) means \(\frac{3}{b^{3}}\) and the \(3\) stays on top.
Practice
Simplify \((5c^{-2})^{-2}\) with no negative exponent left. Type it like \(y^3/8\).
Show the solution
Take the two factors inside separately. The coefficient gives \(5^{-2}=\frac{1}{25}\), and power of a power multiplies exponents so \((c^{-2})^{-2}=c^{4}\), leaving \(\boxed{\frac{c^{4}}{25}}\). Leaving the \(5\) untouched gives \(5c^{4}\), which is 125 times too big.
Practice
Simplify \((3h^{-6})(8h^{2})\) with no negative exponent left. Type it like \(5/y^3\).
Show the solution
Coefficients multiply, \(3\cdot 8=24\), and exponents add, \(-6+2=-4\), so the product is \(24h^{-4}=\boxed{\frac{24}{h^{4}}}\). Only the factor carrying the exponent crosses the bar, so the \(24\) stays on top.
Practice
Simplify \(\frac{30t^{-2}}{5t^{-9}}\) to a single term, then evaluate it at \(t=2\). Enter the value.
Show the solution
The coefficients divide, \(30\div 5=6\), and the exponents subtract, \(-2-(-9)=7\), so the term is \(6t^{7}\). At \(t=2\) that is \(6\cdot 2^{7}=6\cdot 128=\boxed{768}\). Dropping the second minus gives \(-2-9=-11\) and the term \(6t^{-11}\), which at \(t=2\) is \(\frac{3}{1024}\).