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

Mar 3, 2016

$x = 1 + 5 {e}^{- 3}$

#### Explanation:

$\ln \left({x}^{2} - x\right) - \ln \left(5 x\right) = - 3$

Remember that we only can apply logarithms to positive numbers:

So ${x}^{2} - x > 0 \mathmr{and} 5 x > 0$

$x \left(x - 1\right) > 0 \mathmr{and} x > 0 \implies x > 1$

Now, let's solve the equation:

$\ln \left({x}^{2} - x\right) = - 3 + \ln \left(5 x\right)$

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

$\ln \left({x}^{2} - x\right) = \ln \left({e}^{- 3}\right) + \ln \left(5 x\right)$

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

$\ln \left({x}^{2} - x\right) = \ln \left(5 {e}^{- 3} x\right)$

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

${x}^{2} - x = 5 {e}^{- 3} x$

${x}^{2} - \left[5 {e}^{- 3} + 1\right] x = 0$

$\left\{x - \left[5 {e}^{- 3} + 1\right]\right\} x = 0$

$\cancel{x = 0}$(not in dominium) or $x = 1 + 5 {e}^{- 3}$