Solution assignment 06 Logarithmic equations

Solve:

$\ln(2x+1)+\ln(x+2)=\ln(x+10)$

The domains of the subsequent $\ln$ functions are:

$x>-\displaystyle\frac{1}{2}$ and $x>-2$ and $x>-10$

thus:

$x>-\displaystyle\frac{1}{2}$

We rewite the equation as:

$\ln(\displaystyle\frac{(2x+1)(x+2)}{x+10})=0=\ln(1)$

and thus:

$\displaystyle\frac{(2x+1)(x+2)}{x+10}=1$

Cross-multiplication yields:

$(2x+1)(x+2)=x+10$

$2x^2+5x+2=x+10$

$x^2+2x-4=0$

The solutions of this equations are (applying the $abc$ -formula):

$x_{1,2}=-1\pm\sqrt{5}$

The solution:

$x=-1+\sqrt{5}$

lies in the domain and is valid. The other solution lies outside the domain and does not satisfy the equation.