What is x if #ln(x^2-x)-ln(5x) = -3#?

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Answer 1 · 2016-03-03T15:37:03

#x=1+5e^(-3)#

#ln(x^2-x)-ln(5x)=-3#

Remember that we only can apply logarithms to positive numbers:

So #x^2-x>0 and 5x>0#

#x(x-1)>0 and x>0 => x>1#

Now, let's solve the equation:

#ln(x^2-x)=-3+ln(5x)#

#color(red)(a=ln(e^a)#

#ln(x^2-x)=ln(e^(-3))+ln(5x)#

#color(red)(ln(a)+ln(b)=ln(a*b)#

#ln(x^2-x)=ln(5e^(-3)x)#

#color(red)(ln(a)=ln(b)=> a=b#

#x^2-x=5e^(-3)x#

#x^2-[5e^(-3)+1]x=0#

#{x-[5e^(-3)+1]}x=0#

#cancel(x=0)#(not in dominium) or #x=1+5e^(-3)#