x and y are real numbers satisfying the equations below.
{ lo g 1 0 ( 2 x ) + lo g 1 0 y = 2 lo g 1 0 x 2 − lo g 1 0 ( 2 y ) = 4
If x + y = n m , where m and n are positive coprime integers, find m − 3 n 6 .
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{ lo g 1 0 ( 2 x ) + lo g 1 0 y = lo g 1 0 2 + lo g 1 0 x + lo g 1 0 y = 2 lo g 1 0 x 2 − lo g 1 0 ( 2 y ) = 2 lo g 1 0 x − lo g 1 0 2 − lo g 1 0 y = 4 . . . ( 1 ) . . . ( 2 )
From ( 1 ) + ( 2 ) :
3 lo g 1 0 x lo g 1 0 x ⟹ x = 6 = 2 = 1 0 2 = 1 0 0
From ( 1 ) :
lo g 1 0 2 + 2 + lo g 1 0 y lo g 1 0 y ⟹ y = 2 = − lo g 1 0 2 = 2 1
Therefore x + y = 1 0 0 + 2 1 = 2 2 0 1 . Then m − 3 n 6 = 2 0 1 − 3 ( 2 6 ) = 9 .
log 2x+log y=2
2xy=100 ....... (1)
and log x square-log 2y=4
x square/2y=10000 .......(2)
from 1 and 2
x cube=10 power 6
y=1/2
x+y=100+1/2=201/2
m=201 and n=2
m-3n power 6=201-3(2)^6=201-192=9
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Notation: lo g x = lo g 1 0 x
We can use log properties to make both equations look nicer. Let's start with the top one: lo g 2 + lo g y ⟹ lo g 2 x y ⟹ 2 x y = 2 = lo g 1 0 0 = 1 0 0
Now for the second one: lo g x 2 − lo g 2 y ⟹ lo g 2 y x 2 ⟹ 2 y x 2 = 4 = lo g 1 0 4 = 1 0 4
We can rearrange the top one for 2 y : 2 x y ⟹ 2 y = 1 0 0 = x 1 0 0
We can plug that into the second equation and solve for x : 2 y x 2 ⟹ x 1 0 0 x 2 ⟹ 1 0 0 x 3 ⟹ x 3 ⟹ x = 1 0 4 = 1 0 4 = 1 0 4 = 1 0 6 = 1 0 0
We can use this to solve for y : 2 y ⟹ 2 y ⟹ y = x 1 0 0 = 1 0 0 1 0 0 = 1 = 2 1
Now that we have x and y , we can see that x + y = 2 2 0 1 where m = 2 0 1 and n = 2 . Now we can use this to find our answer: m − 3 n 6 = 2 0 1 − 3 ( 2 ) 6 = 2 0 1 − 1 9 2 = 9