Given log(5x+2) = log(2x-5)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)10log5x+2=10log2x−5
Step 2: Simplify, since 10^logA= A10logA=A
5x+2= 2x-55x+2=2x−5
Step 3: Subtract color(red) 22 and color(blue)(2x)2x to both side of the equation to get
5x+2color(red)(-2)color(blue)(-2x)= 2x color(blue)(-2x)-5color(red)(-2)5x+2−2−2x=2x−2x−5−2
3x= -73x=−7
Step 4: Dive both side by 3
(3x)/3 = -7/3 hArr x= -7/33x3=−73⇔x=−73
Step 5: Check the solution
log[(5*-7/3)+2] = log[(2*-7/3) -5]log[(5⋅−73)+2]=log[(2⋅−73)−5]
log(-35/3 + 6/3) = log(-14/3 -15/3)log(−353+63)=log(−143−153)
log(-29/3) = log(-29/3)log(−293)=log(−293)
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>0logbx=y,,x>0,b>0
x= -7/3x=−73 , assuming a complex valued logarithm
