20:[["$","title","0",{"children":"JIPMAT - If 2^x = 3^y = 6^-z, then (1/x + 1/y + 1/z) is equal to | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Let's call the common value a. 2^x = 3^y = 6^-z = a

Converting to logarithmic form: From 2^x = a, we >=t x = _2(a) From 3^y = a, we >=t y = _3(a) From 6^-z = a, we >=t z = -_6(a)

Finding the reciprocals: 1x = 1_2(a) 1y = 1_3(a) 1z = 1-_6(a) = -1_6(a)

Using the chan>= of base formula to standardize: _3(a) = _2(a)_2(3) _6(a) = _2(a)_2(6) This gives: 1y = _2(3)_2(a) 1z = -_2(6)_2(a)

Adding the fractions: 1x + 1y + 1z = 1_2(a) + _2(3)_2(a) - _2(6)_2(a)= 1 + _2(3) - _2(6)_2(a)

Since _2(6) = _2(2 3) = _2(2) + _2(3) = 1 + _2(3): 1 + _2(3) - _2(6)_2(a) = 1 + _2(3) - (1 + _2(3))_2(a) = 0_2(a) = 0

1x + 1y + 1z = 0"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"JIPMAT,QA,Algebra,Indices"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"name":"mobile-web-app-capable","content":"yes"}],["$","meta","8",{"name":"apple-mobile-web-app-title","content":"AfterBoards"}],["$","meta","9",{"name":"apple-mobile-web-app-status-bar-style","content":"default"}],["$","meta","10",{"property":"og:title","content":"JIPMAT - If 2^x = 3^y = 6^-z, then (1/x + 1/y + 1/z) is equal to | PYQs + Solutions"}],["$","meta","11",{"property":"og:description","content":"undefined Let's call the common value a. 2^x = 3^y = 6^-z = a

Converting to logarithmic form: From 2^x = a, we >=t x = _2(a) From 3^y = a, we >=t y = _3(a) From 6^-z = a, we >=t z = -_6(a)

Finding the reciprocals: 1x = 1_2(a) 1y = 1_3(a) 1z = 1-_6(a) = -1_6(a)

Using the chan>= of base formula to standardize: _3(a) = _2(a)_2(3) _6(a) = _2(a)_2(6) This gives: 1y = _2(3)_2(a) 1z = -_2(6)_2(a)

Adding the fractions: 1x + 1y + 1z = 1_2(a) + _2(3)_2(a) - _2(6)_2(a)= 1 + _2(3) - _2(6)_2(a)

Since _2(6) = _2(2 3) = _2(2) + _2(3) = 1 + _2(3): 1 + _2(3) - _2(6)_2(a) = 1 + _2(3) - (1 + _2(3))_2(a) = 0_2(a) = 0

1x + 1y + 1z = 0"}],["$","meta","12",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/jipmat/2021/QA/33"}],["$","meta","13",{"property":"og:site_name","content":"AfterBoards | JIPMAT Preparation"}],["$","meta","14",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","15",{"property":"og:type","content":"website"}],["$","meta","16",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","17",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","18",{"name":"twitter:title","content":"JIPMAT - If 2^x = 3^y = 6^-z, then (1/x + 1/y + 1/z) is equal to | PYQs + Solutions"}],["$","meta","19",{"name":"twitter:description","content":"undefined Let's call the common value a. 2^x = 3^y = 6^-z = a

Converting to logarithmic form: From 2^x = a, we >=t x = _2(a) From 3^y = a, we >=t y = _3(a) From 6^-z = a, we >=t z = -_6(a)

Finding the reciprocals: 1x = 1_2(a) 1y = 1_3(a) 1z = 1-_6(a) = -1_6(a)

Using the chan>= of base formula to standardize: _3(a) = _2(a)_2(3) _6(a) = _2(a)_2(6) This gives: 1y = _2(3)_2(a) 1z = -_2(6)_2(a)

Adding the fractions: 1x + 1y + 1z = 1_2(a) + _2(3)_2(a) - _2(6)_2(a)= 1 + _2(3) - _2(6)_2(a)

Since _2(6) = _2(2 3) = _2(2) + _2(3) = 1 + _2(3): 1 + _2(3) - _2(6)_2(a) = 1 + _2(3) - (1 + _2(3))_2(a) = 0_2(a) = 0

1x + 1y + 1z = 0"}],"$L2e","$L2f","$L30","$L31","$L32","$L33","$L34","$L35","$L36","$L37"]

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