If a > b > 1 and lo g a b + 1 1 6 lo g b a = 1 1 6 7 , what is the value of a 2 + b 1 1 a + b 2 2 ?
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lo g a b + 1 1 6 lo g b a = 1 1 6 7
lo g a b + 1 1 6 ( lo g a b 1 ) = 1 1 6 7
Let x = lo g a b
x + 1 1 x 6 = 1 1 6 7
Multiply both sides by 1 1 x
1 1 x 2 + 6 = 6 7 x
1 1 x 2 − 6 7 x + 6 = 0
( 1 1 x − 1 ) ( x − 6 ) = 0
x = 1 1 1 ; x = 6
x = lo g a b = 1 1 1 ; a 1 1 1 = b ; a = b 1 1
x = lo g a b = 6 ; a 6 = b
Since a is greater than b, we disregard the a 6 = b because both must be greater than 1 .
Solve for a 2 + b 1 1 a + b 2 2 .
( b 1 1 ) 2 + b 1 1 b 1 1 + b 2 2 = b 2 2 + b 1 1 b 1 1 + b 2 2 = 1
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Let lo g a b = x . Then lo g b a = x 1 Rearrange the given information. x + 1 1 x 6 = 1 1 6 7 Multiplying by 1 1 x , 1 1 x 2 + 6 = 6 7 x , so 1 1 x 2 − 6 7 x + 6 = 0 . Solving this equation for x yields that x = 6 or x = 1 1 1 . Since x = lo g a b and a > b , lo g a b < 1 . This causes x = 6 to be an extraneous solution, so x = lo g a b = 1 1 1 . This means that b = 1 1 a . Now plug this in to the expression we are trying to find. a 2 + b 1 1 a + b 2 2 = a 2 + ( 1 1 a ) 1 1 a + ( 1 1 a ) 2 2 = a 2 + a a + a 2 = 1