d:[["$","title","0",{"children":"IIM Kozhikode - Simplify the following: 1_8(x) - 1_7(x) + 1_6(x) - 1_5(x) + 1_4(x) - 1_3(x) + 1_2(x) | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at this expression, I can use the chan>= of base formula to make it much easier! Key Formula: 1_a(x) = _x(a)

Let me rewrite each fraction: 1_8(x) = _x(8) 1_7(x) = _x(7) 1_6(x) = _x(6) 1_5(x) = _x(5) 1_4(x) = _x(4) 1_3(x) = _x(3) 1_2(x) = _x(2)

Now our expression becomes: _x(8) - _x(7) + _x(6) - _x(5) + _x(4) - _x(3) + _x(2)

Using logarithm properties _a(m) + _a(n) = _a(mn) and _a(m) - _a(n) = _a(mn): _x(8 6 4 27 5 3)

Calculating the fraction: Numerator: 8 6 4 2 = 384 Denominator: 7 5 3 = 105 So we have: _x(384105) = _x(12835) (Dividing the numerator and denominator by 3)"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Modern Math,Logarithms"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"IIM Kozhikode - Simplify the following: 1_8(x) - 1_7(x) + 1_6(x) - 1_5(x) + 1_4(x) - 1_3(x) + 1_2(x) | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at this expression, I can use the chan>= of base formula to make it much easier! Key Formula: 1_a(x) = _x(a)

Let me rewrite each fraction: 1_8(x) = _x(8) 1_7(x) = _x(7) 1_6(x) = _x(6) 1_5(x) = _x(5) 1_4(x) = _x(4) 1_3(x) = _x(3) 1_2(x) = _x(2)

Now our expression becomes: _x(8) - _x(7) + _x(6) - _x(5) + _x(4) - _x(3) + _x(2)

Using logarithm properties _a(m) + _a(n) = _a(mn) and _a(m) - _a(n) = _a(mn): _x(8 6 4 27 5 3)

Calculating the fraction: Numerator: 8 6 4 2 = 384 Denominator: 7 5 3 = 105 So we have: _x(384105) = _x(12835) (Dividing the numerator and denominator by 3)"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/28"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | IIM Kozhikode Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"IIM Kozhikode - Simplify the following: 1_8(x) - 1_7(x) + 1_6(x) - 1_5(x) + 1_4(x) - 1_3(x) + 1_2(x) | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at this expression, I can use the chan>= of base formula to make it much easier! Key Formula: 1_a(x) = _x(a)

Let me rewrite each fraction: 1_8(x) = _x(8) 1_7(x) = _x(7) 1_6(x) = _x(6) 1_5(x) = _x(5) 1_4(x) = _x(4) 1_3(x) = _x(3) 1_2(x) = _x(2)

Now our expression becomes: _x(8) - _x(7) + _x(6) - _x(5) + _x(4) - _x(3) + _x(2)

Using logarithm properties _a(m) + _a(n) = _a(mn) and _a(m) - _a(n) = _a(mn): _x(8 6 4 27 5 3)

Calculating the fraction: Numerator: 8 6 4 2 = 384 Denominator: 7 5 3 = 105 So we have: _x(384105) = _x(12835) (Dividing the numerator and denominator by 3)"}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f"]

IIM Kozhikode (QA) PYQs

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