Question

If log 2, log (2x – 1) and log (2x + 3) are in AP then x =

Answer 1

Hey Buddy

Here's The Answer

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Properties used :

=> log a + log b = log ( ab )

=> log a - log b = log ( a/b )

Now, ATQ

log 2, log ( 2x - 1 ) and log ( 2x + 3 ) are in A.P.

So,

=> log ( 2x - 1 ) - log 2 = log ( 2x + 3 ) - log ( 2x - 1 )

Using Property

=> log [ ( 2x - 1 )/2 ] = log [ ( 2x + 3 )/( 2x - 1 ) ]

Cancelling log both side

=> ( 2x - 1 ) / 2 = ( 2x + 3 ) / ( 2x - 1 )

Cross multiplying

=> ( 2x - 1 ) × ( 2x - 1 ) = 2 ( 2x + 3 )

=> 4x² - 4x + 1 = 4x + 6

=> 4x² - 8x - 5 = 0

Factorising

=> 4x² + 2x - 10x - 5 = 0

=> 2x ( 2x + 1 ) - 5 ( 2x + 1 ) = 0

=> ( 2x + 1 ) ( 2x - 5 ) = 0

Case I

=> ( 2x - 5 ) = 0

=> x = 5/2 ✓

Case II

=> ( 2x + 1 ) = 0

=> x = -1/2 ( rejected )

For example, if we try to put ( x = -1/2 ) in log ( 2x - 1 )

=> log ( 2( -1/2 ) - 1 )

The final result will be -ve in log, which is not possible, hence the value of x should be

x = 5/2 ✓

Hope It Helps