If x^2 + y^2 = 34xy, then prove that 2log(x + y) = 2log(6) + log(x) + log(y).

We proved that if x 2 + y 2 = 34 x y , then indeed 2 lo g ( x + y ) = 2 lo g ( 6 ) + lo g ( x ) + lo g ( y ) holds. This was achieved by demonstrating that ( x + y ) 2 = 36 x y . Thus, the logarithmic identity holds true as per the properties of logarithms.
;

Answered by Anonymous | 2025-07-03

To prove that 2 lo g ( X + Y ) = 2 lo g ( 6 ) + lo g ( X ) + lo g ( Y ) , given that X 2 + Y 2 = 34 X Y , we will first simplify the given equation.
Step 1: Simplify X 2 + Y 2 = 34 X Y
The given equation is: X 2 + Y 2 = 34 X Y .
This resembles the identity for the expansion of a square of a sum and difference: X 2 + Y 2 = ( X + Y ) 2 − 2 X Y .
Therefore, let's rearrange the given equation: ( X + Y ) 2 − 2 X Y = 34 X Y .
Solving for ( X + Y ) 2 , we combine like terms: ( X + Y ) 2 = 34 X Y + 2 X Y = 36 X Y .
Step 2: Relate to logarithmic expressions
We are tasked to prove: 2 lo g ( X + Y ) = 2 lo g ( 6 ) + lo g ( X ) + lo g ( Y ) .
Rewriting ( X + Y ) 2 = 36 X Y in logarithmic form: lo g (( X + Y ) 2 ) = lo g ( 36 X Y ) .
This simplifies to: 2 lo g ( X + Y ) = lo g ( 36 ) + lo g ( X Y ) .
Step 3: Further Simplifying
Using the property of logarithms, lo g ( A B ) = lo g ( A ) + lo g ( B ) , we can expand the right side: lo g ( 36 ) + lo g ( X Y ) = lo g ( 36 ) + lo g ( X ) + lo g ( Y ) .
Since 36 = 6 2 , lo g ( 36 ) can be expressed as: lo g ( 6 2 ) = 2 lo g ( 6 ) .
Therefore, combining all, we have: 2 lo g ( X + Y ) = 2 lo g ( 6 ) + lo g ( X ) + lo g ( Y ) .
This completes the proof. Hence, the original statement is proven correct given the initial condition X 2 + Y 2 = 34 X Y .

Answered by LiamAlexanderSmith | 2025-07-07