Question

Solve the equation $\frac{3}{2}{\log }_4(x+2{)}^2+3={\log }_4(4-x{)}^3+{\log }_4(6+x{)}^3$.

Answer 1

$\frac{3}{2}{\log }_4(x+2{)}^2+3={\log }_4(4-x{)}^3+{\log }_4(6+x{)}^3$

$\frac{3}{2}\times 2{\log }_4(x+2)+3{\log }_{4}4=3{\log }_4(4-x)+3{\log }_4(6+x)$ -----since, $\log a^n=n\log a$

$3\left({\log }_4\right(x+2)+{\log }_{4}4)=3\left({\log }_4\right(4-x)+{\log }_4(6+x\left)\right)$

${\log }_4(x+2)+{\log }_{4}4={\log }_4(4-x)+{\log }_4(6+x)$

${\log }_4(x+2)4={\log }_4(4-x)(6+x)$ ----- Product law of logarithms

$(x+2)4=(4-x)(6+x)$

$4x+8=-x^2-2x+24$

$x^2+6x-16=0$

$x^2+8x-2x-16=0$

$(x+8)(x-2)=0$

$x=-8$, $x=2$