Question

log6(x+3)+log6(x+4)=1

Answer 1

First, we must know the condition of solution.
The equation has sense iff
\[x+3>0\text{ et } x+4>0\]
This is equivalent to :
\[x>-3\]

We have :
\[\log_6(x+3)+\log_6(x+4)=1\iff \log_6\left[(x+3)(x+4)\right]=1
\\\iff (x+3)(x+4)=e\\
\iff x^2+3x+4x+12=e\\
\iff x^2+7x+12-e=0

\]
And we have :
\[\Delta=49-4\times(12-e)=1+4e\]
So :
\[x_1=\frac{-7+\sqrt{1+4e}}{2}\simeq -1.78>-3\quad\text{accepted}\\
x_2=\frac{-7-\sqrt{1+4e}}{2}\simeq-5.22<-3 \quad\text{rejected}\]