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

##### 1 Answer

Jan 21, 2016

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#