Add, subtract and multiply indices. Part of HELM Book 1.2
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "There is a set of rules that enables us to manipulate expressions involving indices. These rules are known as the laws of indices, and they occur so commonly that it is worthwhile to memorise them.
\nThe laws of indices state:
\n| First law: | \n$a^m\\times a^n = a^{m+n}$ | \nadd indices when multiplying numbers with the same base | \n
| Second law: | \n$\\frac{a^m}{a^n} = a^{m-n}$ | \nsubtract indices when dividing numbers with the same base | \n
| Third law: | \n$(a^m)^n = a^{mn}$ | \nmultiply indices together when raising a number to a power | \n
\n
Example 16
\nSimplify (a) $a^5\\times a^4$, $\\qquad$ (b) $2x^5(x^3)$.
\nSolution
\nIn each case we are required to multiply expressions involving indices. The bases are the same and we use the first law of indices.
(a) The indices must be added, thus $a^5\\times a^4 = a^{5+4} = a^9$.
(b) Because of the associativity of multiplication we can write
\\(2x^5(x^3)=2(x^5x^3)=2x^{(5+3)}=2x^8\\)
\nThe first law of indices (Key Point 5) extends in an obvious way when more terms are involved:
\nExample 17
\nSimplify $b^5\\times b^4 \\times b^7$.
\nSolution
\nThe indices are added. Thus $b^5\\times b^4 \\times b^7=b^{5+4+7}=b^{16}$.
\n", "advice": "Task 1:
\n$\\var{a_name}^{\\var{i1_expr}}\\times \\var{a_name}^{\\var{i2_expr}}\\times \\var{a_name}^{\\var{i3_expr}} = \\var{a_name}^{\\var{i1_expr}+\\var{i2_expr}+\\var{i3_expr}} = \\var{a_ans}$
\nTask 2:
\n$\\var{b_name}^\\var{b_i1_expr}\\div\\var{b_name}^\\var{b_i2_expr} = \\var{b_ans}$
\nTask 3:
\n$(\\var{c_name}^\\var{c_i1_expr})^\\var{c_i2_expr} = \\var{c_name}^{\\var{c_i1_expr}\\times\\var{c_i2_expr}}=\\var{c_ans}$
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\nNote: to enter $x^{(y + z)}$, type x^(y+z)
All quantities have the same base. To multiply the quantities together, the indices are added.
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\nSimplify (a) $\\displaystyle{\\frac{8^4}{8^2}}$,$\\qquad$(b)$x^{18}\\div x^7$.
\nSolution
\nIn each case we are required to divide expressions involving indices. The bases are the same and we use the second law of indices (Key Point 5).(a) The indices must be subtracted, thus $\\displaystyle{\\frac{8^4}{8^2}=8^{4-2}=8^2=64}$.
(b) Again the indices are subtracted, and so $x^{18}\\div x^7 = x^{18-7}=x^{11}$
Simplify $\\var{b_name}^\\var{b_i1_expr}\\div\\var{b_name}^\\var{b_i2_expr}$.
\nNote: to enter $x^{(y + z)}$, type x^(y+z)
The bases are the same, and the division is carried out by subtracting the indices.
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\nSimplify (a) $(8^2)^3$,$\\qquad$(b)$(z^3)^4$.
\nSolution
\nWe use the third law of indices (Key Point 5).
\n(a) $(8^2)^3 = 8^{2\\times 3} = 8^6$
\n(b) $(z^3)^4 = z^{3\\times 4} = z^{12}$
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\nNote: to enter $x^{yz}$, type x^(yz) or x^(y*z)
Using the third law of indices, the two powers are multiplied.
"}], "answer": "{c_ans}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Two important results which can be derived from the laws of indices state:
\nAny non-zero number raised to the power $0$ has the value $1$, that is $a^0 = 1$.
\nAny number raised to power $1$ is itself, that is $a^1 = a$.
\nA generalisation of the third law states:
\n\\( (a^m b^n)^k = a^{mk}b^{nk} \\)
\n\n
Example 20
\nRemove the brackets from (a) $(3x)^2$,$\\quad$ (b) $(x^3y^7)^4$.
\nSolution
\n(a) Noting that $3=3^1$ and $x=x^1$ then $(3x)^2=(3^1x^1)^2=3^2x^2=9x^2$
\nor, alternatively, $(3x)^2=(3x)\\times(3x)=9x^2$
\n(b) $(x^3y^7)^4=x^{3\\times 4}y^{7\\times 4} = x^{12}y^{28}$
\nShow that $(-xy)^2$ is equivalent to $x^2y^2$ whereas $(-xy)^3$ is equivalent to $-x^3y^3$.
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