Question

How do you solve `ln(x-2) + ln(2x-3)=2lnx`?

Answer 1

x = 6

Explanation:

using the following laws of logs :

`• logx + logy = logxy...............(1)`

`• logx^n = nlogx............(2)`

ln(x - 2 ) + ln(2x - 3 ) = ln(x - 2 )(2x - 3 ).....from (1)

and `2lnx = lnx^2 color(black)(" ...... from (2)")`

`rArr ln(x - 2 ) + ln(2x - 3 ) = 2lnx`

can be written ln(x - 2 )(2x - 3 ) = `lnx^2`

hence (x - 2 )(2x - 3 ) = `x^2`

so (x - 2 )(2x - 3 ) = `x^2`

(distribute the brackets) to get : `2 x^2 - 7x + 6 = x^2`

(collect like terms and equate to 0)

`x^2 - 7x + 6 = 0`

( factorise the quadratic )

(x - 6 )(x - 1 ) = 0 `rArr x = 6 , x = 1`

now x ≠ 1 since ln(x - 2 ) and ln(2x - 3 ) would be ln(-1) and ln(-1)

`rArr x = 6`