Question

Solve the following equations for x:

(i) $7^{2x+3}=1$

(ii) $2^{x+1}=4^{x-3}$

(iii)$2^{5x+3}=8^{x+3}$

(iv) $4^{2x}=\frac{1}{32}$

(v)$4^{x-1}\times (0.5{)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

(vi)$2^{3x-7}=256$

Answer 1

(i) $7^{2x+3}=1$ $\Rightarrow 7^{2x+3}=7^0\therefore 2x+3=0\Rightarrow 2x=-3x=\frac{-3}{2}$

(ii) $2^{x+1}=4^{x-3}$ = $\Rightarrow 2^{x+1}=(2^2{)}^{x-3}\Rightarrow 2^{x+1}=2^{2x-6}Comparing,wegetx+1=2x-61+6=2x-x\Rightarrow x=7\therefore x=7$

(iii)$2^{5x+3}=8^{x+3}$

$\Rightarrow 2^{5x+3}=(2^3{)}^{x+3}\Rightarrow 2^{5x+3}=2^{3x+9}Comparing,weget5x+3=3x+9\Rightarrow 5x-3x=9-3\Rightarrow 2x=6\Rightarrow x=\frac{6}{2}=3\therefore x=3$

(iv) $4^{2x}=\frac{1}{32}$ $\Rightarrow (2^2{)}^{2x}=\frac{1}{2^5}\Rightarrow 2^{4x}=2^{-5}Comparing,weget4x=-5\Rightarrow x=\frac{-5}{4}\therefore x=\frac{-5}{4}$

(v)$4^{x-1}\times (0.5{)}^{3-2x}={\left(\frac{1}{8}\right)}^x$ $\Rightarrow (2^2{)}^{x-1}\times {\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{2^3}\right)}^x\Rightarrow 2^{2x-2}\times 2^{-3+2x}=2^{-3x}\Rightarrow 2^{2x-2-3+2x}=2^{-3x}\Rightarrow 2^{4x-5}=2^{-3x}Comparing,weget4x-5=-3x\Rightarrow 4x+3x=5\Rightarrow 7x=5\Rightarrow x=\frac{5}{7}\therefore x=\frac{5}{7}$

(vi)$2^{3x-7}=256$ = $\Rightarrow 2^{3x-7}=2^{8}Comparing,weget\Rightarrow 3x=8+7=15\Rightarrow x=\frac{15}{3}=5\therefore x=5$