The sum of all the real solutions of the equation _(x+3)(6x^2+28x+30)=5-2 _(6x+10)(x^2+6x+…

Question

The sum of all the real solutions of the equation ${\log }_{(x+3)}(6{x}^{2}+28x+30)=5-2{\log }_{(6x+10)}({x}^{2}+6x+9)$ is equal to :

Accepted Answer

$\begin{gathered} {\log }_{(x+3)}(6{x}^{2}+28x+30)=5-2{\log }_{(6x+10)}({x}^{2}+6x+9)-1 \ {\log }_{(x+3)}2(3{x}^{2}+14x+15)+2{\log }_{(6x+10)}({x}^{2}+6x+9)=5 \ {\log }_{(x+3)}2(x+3)(3x+5)+2{\log }_{(6x+10)}(x+3{)}^{2}=5 \ {\log }_{(x+3)}(x+3)+{\log }_{(x+3)}(6x+10)+4{\log }_{(6x+10)}(x+3)=5 \ {\log }_{(x+3)}(6x+10)+4{\log }_{(6x+10)}(x+3)=4\, {{\log }_{(x+3)}(x+3)=1} \end{gathered}$

Let, ${\log }_{(x+3)}(6x+10)=t$

$\Rightarrow {\log }_{(6x+10)}(x+3)=\frac{1}{t}$

$	herefore $ $t+4	imes \frac{1}{t}=4$

$\begin{gathered} {t}^{2}-4t+4=0 \ (t-2{)}^{2}=0\implies t=2 \end{gathered}$

$\begin{gathered} x=1 \ {\log }_{(1+3)}(6(1{)}^{2}+28(1)+30)=5-2{\log }_{(6(1)+10)}({1}^{2}+6(1)+9) \ {\log }_{4}64=5-2{\log }_{16}16 \ 3=5-2=3 \ x=-1 \ {\log }_{(-1+3)}(6(-1{)}^{2}+28(-1)+30)=5-2{\log }_{(6(-1)+10)}((-1{)}^{2}+6(-1)+9) \ {\log }_{2}8=5-2{\log }_{4}4 \ 3=3 \ x=1x=-1 \ =1+(-1)=0 \end{gathered}$

Solutions · 1