3rd Problem

Algebra Level 3

log ⁡ 2 x 6 + log ⁡ x 2 = 5 , x = ? \large \log_2 x^6 + \log_x 2 = 5, \ \ \ \ \ x = \ ?

2 3 \sqrt[3]{2} 2 \sqrt {2} 2 , 2 3 \sqrt 2, \sqrt[3]{2} No such value exist

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Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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1 solution

log ⁡ 2 x 6 + log ⁡ x 2 = 5 6 log ⁡ 2 x + 1 log ⁡ 2 x = 5 \log _{ 2 }{ { x }^{ 6 } } +\log _{ x }{ 2 } =5\\ 6\log _{ 2 }{ { x } } +\frac { 1 }{ \log _{ 2 }{ x } } =5

multiplying both sides with log ⁡ 2 x \log _{ 2 }{ x } give us

6 ( log ⁡ 2 x ) 2 + 1 = 5 log ⁡ 2 x 6 ( log ⁡ 2 x ) 2 − 5 log ⁡ 2 x + 1 = 0 ( 3 log ⁡ 2 x − 1 ) ( 2 log ⁡ 2 x − 1 ) = 0 log ⁡ 2 x = 1 3 ⇒ x = 2 3 log ⁡ 2 x = 1 2 ⇒ x = 2 6(\log _{ 2 }{ { x } })^2 +1=5\log _{ 2 }{ { x } } \\ 6(\log _{ 2 }{ { x } })^2 -5\log _{ 2 }{ { x } } +1=0\\ \left( 3\log _{ 2 }{ { x }-1 } \right) \left( 2\log _{ 2 }{ { x }-1 } \right) =0\\ \log _{ 2 }{ { x } } =\frac { 1 }{ 3 } \Rightarrow x=\sqrt [ 3 ]{ 2 } \\ \log _{ 2 }{ { x } } =\frac { 1 }{ 2 } \Rightarrow x=\sqrt { 2 }

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sir,how could we change the base of log by taking inverse of log

Krishan Chandra Pandey - 6 years, 1 month ago

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remember this formula: log ⁡ a b = log ⁡ x b log ⁡ x a \log _{ a }{ b } =\frac { \log _{ x }{ b } }{ \log _{ x }{ a } } subtutite x = b x=b we get log ⁡ a b = log ⁡ b b log ⁡ b a = 1 log ⁡ b a \log _{ a }{ b } =\frac { \log _{ b }{ b } }{ \log _{ b }{ a } } =\frac { 1 }{ \log _{ b }{ a } }

Muhammad Fairuzi Teguh - 6 years, 1 month ago

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