Question

If x, y, z are the consecutive positive integers, then log ( 1 + zx ) = ?

i ) Log y
ii ) Log ( y/2 )
iii ) Log ( 2y )
iv ) 2 Log y

Answer with steps...!!

Answer 1

HEY BUDDY!!

HERE'S THE ANSWER.

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♠️ If x , y and z are are the consecutive positive integers ,

▶️ let ( x = n ) , ( y = n + 1 ) and ( z = n + 2 ) // consecutive .

✔️ log [ 1 + x . z ]

=> log [ 1 + ( n ) . ( n + 2 ) ] // by putting values

=> log [ 1 + ( n )^2 + 2 n ]

=> log [ ( n )^2 + 2 n + 1 ]

=> log [ ( n )^2 + n + n + 1 ]

=> log [ n ( n + 1 ) + 1 ( n + 1 ) ]

=> log [ ( n + 1 ) . ( n + 1 ) ]

=> log [ ( n + 1 )^2 ]_______( 1 )

⏺️ Now by putting value of ( n + 1 ) in expression ( 1 )

=> log [ ( y )^2 ]

▶️ Using property of log , { log n^m = m log n }

=> [ 2 log y ] ✔️✔️


♠️ log ( 1 + z x ) = 2 log y , Hence option [ i v ] is correct.


HOPE HELPED..

JAI HIND..

:-)

Answer 2

Hey there!

Given, x, y, z are three consecutive numbers

Let y = a
then , x = a - 1 , z = a + 1

Now,
xz = ( a + 1 ) ( a - 1 ) = a² - 1

log ( 1 + zx )

= log ( 1 + a² - 1)

= log (a²)

= log(y²)

= 2logy

Therefore, The correct option is IV.

Hope helped! ^^