For a , b ∈ R + and > 1 ,
a lo g a b − b lo g b a = ?
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i dont get the part where root loga b became the inverse of root logb a
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In general, the change of base rule gives us that lo g a ( x ) = lo g b ( a ) lo g b ( x ) , where a , b , x are all positive reals.
If x = b then since lo g b ( b ) = 1 we find that lo g a ( b ) = lo g b ( a ) 1 .
Now if both a , b are greater than 1 then both lo g a ( b ) and lo g b ( a ) exceed 0 , which allows us to take the square roots of both sides of this last equation, yielding
lo g a ( b ) = lo g b ( a ) 1 .
(Note that for this question to make any sense it must be assumed that a , b > 1 , so while this condition is not explicitly stated it seems reasonable to assume it implicitly.)
First we will focus on a lo g a b , which can be expessed as a ln a ln b according to the change of base rule.
Rationalizing the denominator, we get:
a ln a ln b = a ln a ln b ln a which can also be expresed as ( a ln a 1 ) ln a ln b
Now, by the change of base rule, we have:
ln a = lo g a e lo g a a = lo g a e 1 ⇒ ln a 1 = lo g a e
Hence ( a ln a 1 ) ln a ln b = ( a lo g a e ) ln a ln b = e ln a ln b
A similar simplification holds for b lo g b a , so the original expresion reduces to
e ln a ln b − e ln a ln b = 0
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Suppose x = a lo g a ( b ) . Then by the change of base rule
lo g a ( x ) = lo g a ( b ) ⟹ lo g b ( a ) lo g b ( x ) = lo g b ( a ) 1
⟹ lo g b ( x ) = lo g b ( a ) ⟹ x = b lo g b ( a ) .
The given expression is then equal to x − x = 0 .