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

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