QuanticaAlgebra IOpen in the course1.5 built the exponent laws out of counting, and every one of those counts still works when the base is a letter. A letter stands for a number you have not named yet, so \(k^5\) is five factors of \(k\) and nothing in the argument changes. What is new is knowing when the laws do not apply, and there are two cases, powers with different bases and terms joined by a plus sign instead of multiplied.
Problem
\(m^3\cdot m\cdot m^4\) is a row of factors of \(m\) with nothing but multiplication in it. Written as a single power it is \(m^k\). Count the factors and enter \(k\).
Show a hint
- Write out the factors of \(m\) in each piece, then count the whole row.
- A letter with no exponent written is one factor.
Show the full solution
A letter with no exponent written is one factor, since \(m=m^1\), so the row is 3 factors then 1 then 4. Joining rows adds the counts, giving \(\boxed{8}\). Skipping the bare \(m\) is the easy miss and gives 7.
Problem
A student writes \(n^2+n^5=n^7\), adding the exponents. Test the claim by picking a number, the way 1.4 did. Evaluate \(n^2+n^5\) at \(n=2\) and enter the value.
Show a hint
- \(n^2\) and \(n^5\) are separate terms, so work out each one and then add.
- For comparison, \(n^7\) at \(n=2\) is 128.
Show the full solution
At \(n=2\), \(n^2=4\) and \(n^5=32\), so \(n^2+n^5=4+32=\boxed{36}\). The claim predicts \(n^7=128\), and one mismatch kills it. The product rule applies to a product, and this is a sum.
Problem
Multiply \((4c^5)(7c^3)\) into a single term. The numbers out front and the powers of \(c\) do not get the same treatment, so decide what happens to each before you write anything. Enter the result, typed like \(6y^2\).
Show a hint
- Gather the numbers together and the \(c\) factors together, the 1.3 way.
- The 4 and the 7 are factors. The 5 and the 3 are counts.
Show the full solution
Coefficients are factors, so they multiply, \(4\cdot 7=28\). Exponents are counts, so they add, \(5+3=8\). The term is \(\boxed{28c^8}\). Treating both halves alike gives \(28c^{15}\) or \(11c^8\), each applying the wrong operation to half the term.
Problem
In \(\frac{z^{11}}{z^{4}}\), every factor on the bottom cancels a matching factor on top. What survives is a single power \(z^k\). Enter \(k\).
Show a hint
- Four factors cancel four factors. How many of the eleven are left?
Show the full solution
The four factors on the bottom cancel four of the eleven on top, so \(11-4=\boxed{7}\) factors survive. The 4 is the count that got cancelled, not the count left over.
Problem
Simplify \(\frac{54k^{10}}{6k^{3}}\) to a single term. The number part and the \(k\) part are handled separately. Enter the coefficient of the result.
Show a hint
- Split it into a number part and a \(k\) part.
- The coefficients divide. Only the exponents subtract.
Show the full solution
The coefficients divide, \(54\div 6=9\), and the exponents subtract, \(10-3=7\), so the term is \(9k^{7}\) and the coefficient is \(\boxed{9}\). Subtraction belongs to the exponents. Subtracting the numbers out front would give 48.
Problem
\((r^4)^6\) means six copies of \(r^4\) multiplied together. Written as one power it is \(r^k\). Count the factors of \(r\) in that whole product and enter \(k\).
Show a hint
- Each copy of \(r^4\) contributes four factors of \(r\).
Show the full solution
Each of the six copies of \(r^4\) brings four factors of \(r\), so the total is \(4\cdot 6=\boxed{24}\). The 6 counts copies of a four-factor block, not extra factors, so adding to get 10 uses the product rule where it does not apply.
Problem
Both \(q^5\cdot q^4\) and \((q^5)^4\) collapse to a single power of \(q\), and they do not collapse the same way. Write each one as a single power of \(q\). Take the exponent on \(q\) in the collapsed form of \((q^5)^4\) and subtract the exponent on \(q\) in the collapsed form of \(q^5\cdot q^4\). Enter that one number.
Show a hint
- One expression sets two rows side by side. The other makes four copies of one row.
- \(5+4\) and \(5\cdot 4\) are not the same number.
Show the full solution
Side by side adds the counts, so \(q^5\cdot q^4=q^9\). Four copies multiply them, so \((q^5)^4=q^{20}\). The difference is \(20-9=\boxed{11}\). Reading the two as the same thing gives 0, which is exactly the confusion.
Problem
Expand \((3b^7)^3\) into a single term. The outer exponent makes three copies of everything inside the parentheses, the 3 in front included. Enter the coefficient of the result.
Show a hint
- Write the three copies out. What sits inside each one?
Show the full solution
Three copies of \(3b^7\) hold three 3s and twenty-one \(b\)s, so the term is \(27b^{21}\) and the coefficient is \(\boxed{27}\). The outer exponent reaches the coefficient too, so \(3b^{21}\) keeps only one of those three 3s.
Problem
Expand \((u^3w^8)^4\). Each base carries its own exponent through the expansion. Enter the exponent on \(w\) in the result.
Show a hint
- Four copies of \(u^3w^8\) reorder into all the \(u\) factors and then all the \(w\) factors.
Show the full solution
Every copy of \(u^3w^8\) contains eight \(w\) factors, so four copies give \(u^{12}w^{32}\) and the exponent on \(w\) is \(\boxed{32}\). The outer 4 reaches both bases. Applying it to the first one only leaves \(w^8\) untouched.
Problem
Simplify \((2y^4)^3\cdot 5y^2\) to a single term. Clear the parentheses first, then multiply. Enter the result, typed like \(6k^5\).
Show a hint
- The cube reaches the 2 as well as the \(y^4\).
- Once the parentheses are gone it is one product of two monomials.
Show the full solution
The cube makes three copies of \(2y^4\), so the parentheses clear to \(8y^{12}\). Then \(8\cdot 5=40\) and \(12+2=14\), giving \(\boxed{40y^{14}}\). Skipping the \(2^3\) gives \(10y^{14}\), which is the whole trap.
Problem
Simplify \(\frac{(5f^4)^2\cdot f^3}{5f^{6}}\) to a single term, then evaluate that term at \(f=2\). Enter the value, a single number.
Show a hint
- Clear the parentheses first, then build one term on top before you divide.
- Check what the outside exponent does to the 5, not just to \(f^4\).
Show the full solution
\((5f^4)^2=25f^8\), so the top is \(25f^8\cdot f^3=25f^{11}\). Dividing by \(5f^6\) gives \(5f^5\), and at \(f=2\) that is \(5\cdot 32=\boxed{160}\). The outside exponent reaches the 5 as well as the \(f^4\), so \(5f^8\) is not enough.
Every exponent in this lesson was a positive whole number, and we kept it that way by putting the larger exponent on top of every quotient. 3.2 drops that rule. Once the bottom exponent is allowed to be the larger one, \(x^0\) and \(x^{-3}\) need a meaning, and the meaning is whatever keeps these laws working.
Practice these ideas
Practice
Write \(g^7\cdot g^8\) as a single power \(g^k\) and enter \(k\).
Show the solution
Joining two rows of \(g\) factors adds the counts, \(7+8=\boxed{15}\). Multiplying the exponents would give \(g^{56}\), which counts copies that are not there.
Practice
A student claims \(d^2+d^4=d^6\). Test it the 1.4 way. Evaluate \(d^2+d^4\) at \(d=3\) and enter the value.
Show the solution
At \(d=3\), \(d^2=9\) and \(d^4=81\), so the value is \(9+81=\boxed{90}\). The claim predicts \(3^6=729\), so it fails. A plus sign joins terms, not rows of factors.
Practice
A student claims \(a^4b^2=(ab)^6\). Test the claim by evaluating \(a^4b^2\) at \(a=2\) and \(b=3\), and enter the value.
Show the solution
At \(a=2\) and \(b=3\), \(a^4b^2=16\cdot 9=\boxed{144}\). The claim would give \(6^6=46656\), so it fails. Adding exponents needs a shared base, and \(a\) and \(b\) are different bases.
Practice
Multiply \((7h^2)(2h^6)\) into a single term and enter its coefficient.
Show the solution
\(7\cdot 2=14\), so the term is \(14h^8\) and the coefficient is \(\boxed{14}\). The exponents have nothing to do with the coefficient.
Practice
Multiply \((8p^3)(3p^4)\) into a single term. Enter the result, typed like \(5y^2\).
Show the solution
Coefficients multiply, \(8\cdot 3=24\), while exponents add, \(3+4=7\), giving \(\boxed{24p^7}\). Adding the coefficients to get \(11p^7\) is the tempting slip, but they are factors, not terms.
Practice
Simplify \(\frac{v^{16}}{v^{6}}\) to a single power \(v^k\) and enter \(k\).
Show the solution
Six factors on the bottom cancel six of the sixteen on top, so \(16-6=\boxed{10}\) factors of \(v\) are still standing. Answering 6 reports the count that cancels, not the exponent left behind.
Practice
Simplify \(\frac{48t^{9}}{6t^{4}}\) to a single term. Enter it, typed like \(3y^2\).
Show the solution
The coefficients divide, \(48\div 6=8\), and the counts subtract, \(9-4=5\), giving \(\boxed{8t^5}\). Carrying the 48 along and only subtracting the exponents leaves the number part untouched.
Practice
\((s^6)^7\) is seven copies of \(s^6\) multiplied together. Write it as \(s^k\) and enter \(k\).
Show the solution
Each of the seven copies contributes six factors of \(s\), so the row holds \(6\cdot 7=\boxed{42}\) factors. Adding the exponents gives 13, but that is the product rule, and this is a power raised to a power.
Practice
Expand \((5n^3)^3\) into a single term and enter its coefficient.
Show the solution
The outer exponent applies to every factor inside the parentheses, so \((5n^3)^3=5^3\cdot n^{3\cdot 3}=125n^9\) and the coefficient is \(\boxed{125}\). Leaving the 5 alone gives \(5n^9\) and drops two of the three 5s.
Practice
Simplify each of \((c^3)^8\) and \(c^3\cdot c^8\) to a single power of \(c\). Enter the exponent of the first minus the exponent of the second.
Show the solution
Stacking eight copies of \(c^3\) multiplies the counts, so \((c^3)^8=c^{24}\). Setting two rows side by side adds them, so \(c^3\cdot c^8=c^{11}\). The difference is \(24-11=\boxed{13}\). Reading both the same way gives 0.
Practice
Expand \((j^5z^3)^3\) and enter the exponent on \(z\).
Show the solution
Each of the three copies carries three \(z\) factors, so the result is \(j^{15}z^{9}\) and the exponent on \(z\) is \(\boxed{9}\). The 3 outside reaches both letters, not just the first one.
Practice
Simplify \(\frac{(2m^5)^4}{8m^{11}}\) to a single term. Enter it, typed like \(7v^3\).
Show the solution
The parentheses clear to \(16m^{20}\), then \(16\div 8=2\) and \(20-11=9\), giving \(\boxed{2m^9}\). Forgetting \(2^4\) leaves \(2m^{20}\) on top and wrecks the coefficient.