How do you solve #log(5x+2)=log(2x-5)#?

1 Answer
Dec 9, 2015

#x= -7/3#

Explanation:

Given #log(5x+2) = log(2x-5)# common log- base 10

Step 1: Raised it to exponent using the base 10

#10^(log5x+2) = 10^(log2x-5)#

Step 2: Simplify, since #10^logA= A#
#5x+2= 2x-5#

Step 3: Subtract #color(red) 2# and #color(blue)(2x)# to both side of the equation to get
#5x+2color(red)(-2)color(blue)(-2x)= 2x color(blue)(-2x)-5color(red)(-2)#
#3x= -7#

Step 4: Dive both side by 3
#(3x)/3 = -7/3 hArr x= -7/3#

Step 5: Check the solution

#log[(5*-7/3)+2] = log[(2*-7/3) -5]#
#log(-35/3 + 6/3) = log(-14/3 -15/3)#
#log(-29/3) = log(-29/3)#

Both side are equal, despite we can't take a log of a negative number due to domain restriction #log_b x = y, , x >0 , b>0#
#x= -7/3# , assuming a complex valued logarithm