Question

Solve the following equations.
$3^{4x+8}-{4.3}^{2x+5}+28=2lo{g}_2\sqrt{2}$

Answer 1

$3^{4x+8}-{4.3}^{2x+5}+28=2lo{g}_2\sqrt{2}$

$3^{4x}{.3}^8-{4.3}^{2x}{.3}^5+28=2lo{g}_2{2}^{1/2}=2.\frac{1}{2}lo{g}_{2}2=1$

$\therefore \{lo{g}_a{b}^c=clo{g}_a^b,3^{a+b}=3^a{.3}^b\}$

$3^8.(3^{2x}{)}^2-{4.3}^5{.3}^{2x}+28=1$

$put{3}^{2x}=y$

$3^8.y^2-{4.3}^5.y+27=0$

$y=\frac{{4.3}^5\pm \sqrt{{16.3}^{10}-{4.3}^3{.3}^8}}{{2.3}^8}$

$y=\frac{{4.3}^5\pm 3^5.2\sqrt{4.3-3}}{{2.3}^8}$

$y=\frac{4\pm 2\sqrt{1}}{{2.3}^3}$

$y=\frac{6}{{2.3}^3};y=\frac{2}{{2.}^3}$

$y=\frac{1}{3^2};y=\frac{1}{3^3}$

$\therefore \frac{2x}{3}=3^{-2};3^{2x}=3^{-3}$

$x=-1$

$x=-3/2$