If a , b , c > 0 and x > 1 , let
⎩ ⎪ ⎨ ⎪ ⎧ lo g a b x = 9 lo g b c x = 1 8 lo g a b c x = 8
Then what is the value of lo g a c x = ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Thank you.
⎩ ⎪ ⎨ ⎪ ⎧ lo g a b x = 9 lo g b c x = 1 8 lo g a b c x = 8 ⟹ x = ( a b ) 9 ⟹ x = ( b c ) 1 8 ⟹ x = ( a b c ) 8 ⟹ a b = x 9 1 ⟹ b c = x 1 8 1 ⟹ a b c = x 8 1 . . . ( 1 ) . . . ( 2 ) . . . ( 3 )
⟹ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ( 2 ) ( 3 ) : ( 1 ) ( 3 ) : a = x 7 2 5 c = x 7 2 1 ⟹ a c = x 1 2 1 ⟹ x = ( a c ) 1 2 ⟹ lo g a c x = 1 2
Thank you Sir.
Thanks for sharing your solution.
Problem Loading...
Note Loading...
Set Loading...
Let y = lo g a c x .
Using the logarithmic property, lo g p q = lo g q p 1 , we can rewrite all the given equations as ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ lo g x a b = 9 1 lo g x b c = 1 8 1 lo g x a c = y 1 lo g x a b c = 8 1 .
And because lo g p q + lo g p r = lo g p q r , the sum of the first 3 given equations is equal to twice the remaining given equation, lo g x a b + lo g x b c + lo g x a c = 2 lo g x a b c ⇔ 9 1 + 1 8 1 + y 1 = 8 2 ⇔ y = 1 2 .